Chapter 1 Functions.

Chapter 1 Functions

Chapter Outline The Slope of a Straight Line
The Slope of a Curve at a Point The Derivative Limits and the Derivative Differentiability and Continuity Some Rules for Differentiation More About Derivatives The Derivative as a Rate of Change

§ 1.1 The Slope of a Straight Line

Section Outline Nonvertical Lines
Positive and Negative Slopes of Lines Interpretation of a Graph Properties of the Slope of a Nonvertical Line Finding the Slope and y-Intercept of a Line Sketching the Graph of a Line Making Equations of Lines Slope as a Rate of Change

Nonvertical Lines Definition Example
Equations of Nonvertical Lines: A nonvertical line L has an equation of the form The number m is called the slope of L and the point (0, b) is called the y-intercept. The equation above is called the slope-intercept equation of L. For this line, m = 3 and b = -4.

Lines – Positive Slope EXAMPLE
The following are graphs of equations of lines that have positive slopes.

Lines – Negative Slope EXAMPLE
The following are graphs of equations of lines that have negative slopes.

Interpretation of a Graph
EXAMPLE A salesperson’s weekly pay depends on the volume of sales. If she sells x units of goods, then her pay is y = 5x + 60 dollars. Give an interpretation of the slope and the y-intercept of this straight line. SOLUTION First, let’s graph the line to help us understand the exercise. (80, 460)

Interpretation of a Graph
CONTINUED The slope is 5, or 5/1. Since the numerator of this fraction represents the amount of change in her pay relative to the amount of change in her sales, the denominator, for every 1 sale that she makes, her pay increases by 5 dollars. The y-intercept is 60 and occurs on the graph at the point (0, 60). This point suggests that when she has executed 0 sales, her pay is 60 dollars. This $60 could be referred to as her base pay.

Properties of the Slope of a Nonvertical Line

Properties of the Slope of a Line

Finding Slope and y-intercept of a Line
EXAMPLE Find the slope and y-intercept of the line SOLUTION First, we write the equation in slope-intercept form. This is the given equation. Divide both terms of the numerator of the right side by 3. Rewrite Since the number being multiplied by x is 1/3, 1/3 is the slope of the line. Since the other 1/3 is the number being added to the term containing x, 1/3, or (0, 1/3), is the y-intercept. Incidentally, it was a complete coincidence that the slope and y-intercept were the same number. This does not normally occur.

Sketching Graphs of Lines
EXAMPLE Sketch the graph of the line passing through (-1, 1) with slope ½. SOLUTION We use Slope Property 1. We begin at the given point (-1, 1) and from there, move up one unit and to the right two units to find another point on the line. (-1, 1) (-1, 1)

Sketching Graphs of Lines
CONTINUED Now we connect the two points that have already been determined, since two points determine a straight line.

Making Equations of Lines
EXAMPLE Find an equation of the line that passes through the points (-1/2, 0) and (1, 2). SOLUTION To find an equation of the line that passes through those two points, we need a point (we already have two) and a slope. We do not yet have the slope so we must find it. Using the two points we will determine the slope by using Slope Property 2. We now use Slope Property 3 to find an equation of the line. To use this property we need the slope and a point. We can use either of the two points that were initially provided. We’ll use the second (the first would work just as well).

CONTINUED This is the equation from Property 3. (x1, y1) = (1, 2) and m = 4/3. Distribute. Add 2 to both sides of the equation. NOTE: Technically, we could have stopped when the equation looked like since it is an equation that represents the line and is equivalent to our final equation.

EXAMPLE Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line y = 2x. SOLUTION To find an equation of the line, we need a point (we already have one) and a slope. We do not yet have the slope so we must find it. We know that the line we desire is perpendicular to the line y = 2x. Using Slope Property 5, we know that the product of the slope of the line desired and the slope of the line y = 2x is -1. We recognize that the line y = 2x is in slope-intercept form and therefore the slope of the line is 2. We can now find the slope of the line that we desire. Let the slope of the new line be m. (slope of a line)(slope of a new line) = -1 This is Property 5. 2m = -1 The slope of one line is 2 and the slope of the desired line is denoted by m. m = -0.5 Divide. Now we can find the equation of the desired line using Property 3.

CONTINUED This is the equation from Property 3. (x1, y1) = (2, 0) and m = -0.5. Distribute.

Slope as a Rate of Change
EXAMPLE Compute the rate of change of the function over the given intervals. SOLUTION We first get y by itself in the given equation. This is the given equation. Subtract 2x from both sides. Since this is clearly a linear function (since it’s now in slope-intercept form) it has constant slope, namely -2. Therefore, by definition, it also has a constant rate of change, -2. Therefore, no matter what interval is considered for this function, the rate of change will be -2. Therefore the answer, for both intervals, is -2.

§ 1.2 The Slope of a Curve at a Point

Section Outline Tangent Lines Slopes of Curves
Slope of a Curve as a Rate of Change Interpreting the Slope of a Graph Finding the Equation and Slope of the Tangent Line of a Curve

Tangent Lines Definition Example
Tangent Line to a Circle at a Point P: The straight line that touches the circle at just the one point P

Slope of a Curve & Tangent Lines
Definition Example The Slope of a Curve at a Point P: The slope of the tangent line to the curve at P (Enlargements)

Slope of a Graph EXAMPLE SOLUTION
Estimate the slope of the curve at the designated point P. SOLUTION The slope of a graph at a point is by definition the slope of the tangent line at that point. The figure above shows that the tangent line at P rises one unit for each unit change in x. Thus the slope of the tangent line at P is

Slope of a Curve: Rate of Change

Interpreting Slope of a Graph
EXAMPLE Refer to the figure below to decide whether the following statements about the debt per capita are correct or not. Justify your answers . The debt per capita rose at a faster rate in 1980 than in 2000. The debt per capita was almost constant up until the mid-1970s and then rose at an almost constant rate from the mid-1970s to the mid-1980s.

CONTINUED SOLUTION (a) The slope of the graph in 1980 is marked in red and the slope of the graph in 2000 is marked in blue, using tangent lines. It appears that the slope of the red line is the steeper of the two. Therefore, it is true that the debt per capita rose at a faster rate in 1980.

CONTINUED (b) Since the graph is a straight, nearly horizontal line from 1950 until the mid-1970s, marked in red, it is therefore true that the debt per capita was almost constant until the mid-1970s. Further, since the graph is a nearly straight line from the mid-1970s to the mid-1980s, marked in blue, it is therefore true that the debt per capita rose at an almost constant rate during those years.

Equation & Slope of a Tangent Line
EXAMPLE Find the slope of the tangent line to the graph of y = x2 at the point (-0.4, 0.16) and then write the corresponding equation of the tangent line. SOLUTION The slope of the graph of y = x2 at the point (x, y) is 2x. The x-coordinate of (-0.4, 0.16) is -0.4, so the slope of y = x2 at this point is 2(-0.4) = -0.8. We shall write the equation of the tangent line in point-slope form. The point is (-0.4, 0.16) and the slope (which we just found) is Hence the equation is:

§ 1.3 The Derivative

Section Outline The Derivative Differentiation
Slope and the Derivative Equation of the Tangent Line to the Graph of y = f (x) at (a, f (a)) Leibniz Notation for Derivatives Calculating Derivatives Via the Difference Quotient Differentiable Limit Definition of the Derivative Limit Calculation of the Derivative

The Derivative Definition Example
Derivative: The slope formula for a function y = f (x), denoted: Given the function f (x) = x3, the derivative is

Differentiation Definition Example
Differentiation: The process of computing a derivative. No example will be given at this time since we do not yet know how to compute derivatives. But don’t worry, you’ll soon be able to do basic differentiation in your sleep.

Differentiation Examples
These examples can be summarized by the following rule.

Find the derivative of SOLUTION This is the given equation. Rewrite the denominator as an exponent. Rewrite with a negative exponent. What we’ve done so far has been done for the sole purpose of rewriting the function in the form of f (x) = xr. Use the Power Rule where r = -1/7 and then simplify.

Find the slope of the curve y = x5 at x = -2. SOLUTION We must first find the derivative of the given function. This is the given function. Use the Power Rule. Since the derivative function yields information about the slope of the original function, we can now use to determine the slope of the original function at x = -2. Replace x with -2. Evaluate. Therefore, the slope of the original function at x = -2 is 80 (or 80/1).

Equation of the Tangent Line to the Graph of y = f (x) at the point (a, f (a))

Equation of the Tangent Line
EXAMPLE Find the equation of the tangent line to the graph of f (x) = 3x at x = 4. SOLUTION We must first find the derivative of the given function. This is the given function. Differentiate. Notice that in this case the derivative function is a constant function, 3. Therefore, at x = 4, or any other value, the value of the derivative will be 3. So now we use the Equation of the Tangent Line that we just saw. This is the Equation of the Tangent Line. f (4) = 12 and Simplify.

Leibniz Notation for Derivatives
Ultimately, this notation is a better and more effective notation for working with derivatives.

Calculating Derivatives Via the Difference Quotient
The Difference Quotient is

EXAMPLE Apply the three-step method to compute the derivative of the following function: SOLUTION STEP 1: We calculate the difference quotient and simplify as much as possible. This is the difference quotient. Evaluate f (x + h) and f (x). Simplify. Simplify. Simplify.

CONTINUED Factor. Cancel and simplify. STEP 2: As h approaches zero, the expression -2x – h approaches -2x, and hence the difference quotient approaches -2x. STEP 3: Since the difference quotient approaches the derivative of f (x) = -x2 + 2 as h approaches zero, we conclude that

Differentiable Definition Example
Differentiable: A function f is differentiable at x if approaches some number as h approaches zero. The function f (x) = |x| is differentiable for all values of x except x = 0 since the graph of the function has no definite slope when x = 0 (f is nondifferentiable at x = 0) but does have a definite slope (1 or -1) for every other value of x.

Limit Definition of the Derivative

Limit Calculation of the Derivative
EXAMPLE Using limits, apply the three-step method to compute the derivative of the following function: SOLUTION STEP 1: This is the difference quotient. Evaluate f (x + h) and f (x). Simplify. Simplify. Simplify.

Limit Calculation of the Derivative
CONTINUED Factor. Cancel and simplify. STEP 2: As h approaches 0, the expression -2x – h approaches -2x STEP 3: Since the difference quotient approaches , we conclude that

§ 1.4 Limits and the Derivative

Section Outline Definition of the Limit Finding Limits Limit Theorems
Using Limits to Calculate a Derivative Limits as x Increases Without Bound

Definition of the Limit

Finding Limits EXAMPLE Determine whether the limit exists. If it does, compute it. SOLUTION Let us make a table of values of x approaching 4 and the corresponding values of x3 – 7. x x3 - 7 As x approaches 4, it appears that x3 – 7 approaches 57. In terms of our notation,

Finding Limits EXAMPLE SOLUTION
For the following function g(x), determine whether or not exists. If so, give the limit. SOLUTION We can see that as x gets closer and closer to 3, the values of g(x) get closer and closer to 2. This is true for values of x to both the right and the left of 3.

Limit Theorems

Finding Limits EXAMPLE SOLUTION
Use the limit theorems to compute the following limit. SOLUTION Limit Theorem VI Limit Theorem II with r = ½ Limit Theorem IV

Finding Limits CONTINUED Limit Theorems I and II Since , we have that:

Limit Theorems

Finding Limits EXAMPLE SOLUTION Compute the following limit.
Since evaluating the denominator of the given function at x = 9 is 8 – 3(9) = -19 ≠ 0, we may use Limit Theorem VIII.

Using Limits to Calculate a Derivative

EXAMPLE Use limits to compute the derivative for the function SOLUTION We must calculate

CONTINUED Now that replacing h with 0 will not cause the denominator to be equal to 0, we use Limit Theorem VIII.

More Work With Derivatives and Limits
EXAMPLE Match the limit with a derivative. Then find the limit by computing the derivative. SOLUTION The idea here is to identify the given limit as a derivative given by for a specific choice of f and x. Toward this end, let us rewrite the limit as follows. Now go back to Take f (x) = 1/x and evaluate according to the limit definition of the derivative:

More Work With Derivatives and Limits
CONTINUED On the right side we have the desired limit; while on the left side can be computed using the power rule (where r = -1): Hence,

Limits as x Increases Without Bound
EXAMPLE Calculate the following limit. SOLUTION Both 10x and x2 – 30 increase without bound as x does. To determine the limit of their quotient, we employ an algebraic trick. Divide both numerator and denominator by x2 (since the highest power of x in either the numerator or the denominator is 2) to obtain As x increases without bound, 10/x approaches 0, 100/x2 approaches 0, and 30/x2 approaches 0. Therefore, as x increases without bound, 10/x + 100/x2 approaches = 0 and /x2 approaches 1 – 0 = 1. Therefore,

Limits as x Increases Without Bound
CONTINUED

§ 1.5 Differentiability and Continuity

Section Outline Differentiability and Nondifferentiability
Graphs Nondifferentiable at a Point Continuous and Discontinuous Functions

Examples of Nondifferentiability

Graphs Nondifferentiable at a Point
EXAMPLE The tax that you pay to the federal government is a percentage of your taxable income, which is what remains of your gross income after you subtract your allowed deductions. In the year 2001, there were five rates or brackets for a single taxpayer, as shown below. So if you are single and your taxable income was less than $27,050 in 2001, then your tax is your taxable income times 15% (.15). The maximum amount of tax that you will pay on your income in this first bracket is 15% of $27,050 or (.15) x 27,050 = dollars. If your taxable income is more than $27,050 but less than $65,550, then your tax is $ plus 27.5% of the amount in excess of $27,050. So, for example, if your taxable income is $50,000, then your tax is $ (50,000 – 27,050) = x 22,950 = 10, dollars. Let x denote your taxable income in 2001 and T(x) your tax for that year.

CONTINUED Find a formula for T(x) if x is not over $136,750. Plot the graph of T(x) for 0 ≤ x ≤ 136,750. SOLUTION (a) The function would be:

CONTINUED (b) The graph of T(x) is shown below. Notice that the graph has a corner when x = 27,050 and is therefore nondifferentiable at that point.

Continuous Functions Definition Example
Continuous: A function f is continuous at x = a provided that, roughly speaking, its graph has no breaks (or gaps) as it passes through the point (a, f (a)). That is, f (x) is continuous at x = a provided that we can draw the graph through (a, f (a)) without lifting our pencil from the paper. Notice that f (x) is not continuous at x = 3 but it is continuous for every other value of x.

Differentiability & Continuity

Limits & Continuity In order for (1) to hold, three conditions must be satisfied. f (x) must be defined at x = a. must exist. The limit must have the value f (a). A function will fail to be continuous at x = a when any one of these conditions fails to hold.

EXAMPLE Determine whether the following function is continuous and/or differentiable at x = 1. SOLUTION To save ourselves from doing unnecessary work, let’s check to see if the function is differentiable at x = 1. If it is, Theorem I says that it is automatically continuous at that point (in which case we would not even need to check for continuity). Both pieces of the graph, f (x) = x + 2 and f (x) = 3x meet at x = 1, as seen to the right.

CONTINUED Upon looking at the graph, it appears that the function has a corner at x = 1 and is therefore nondifferentiable at x = 1. However, just looking at a graph is generally not the best way to make an informed decision. If we analyze the function itself at x = 1, namely the slope of the function to either side of x = 1, we find that to the left of x = 1, the slope of the function is 1 (f (x) = x + 2). To the right of x = 1, the slope of the function is 3 (f (x) = 3x). Therefore, in terms of the slope, there is not a smooth transition from the left-most piece of the graph to the right-most piece of the graph. Therefore, the function is not differentiable at x = 1. So what does that say about the continuity of the function at x = 1? Nothing! Theorem I only says that if the function is differentiable at a point then it is continuous at that same point. It says nothing about nondifferentiability implying anything about continuity. Therefore, we must conduct an investigation for the purpose of determining whether the function is continuous at x = 1 or not. Using the Limit Definition of Continuity, we will determine whether the function is continuous at x = 1 or not. We will investigate the three conditions associated with the definition.

CONTINUED Condition 1: The function is defined at x = 1 since f (1) = (1) + 2 = 3. Condition 2: We will show that the limit of the function at x = 1 exists. x x x x + 2 As x approaches 1, it appears that the function approaches 3 from both sides of x = 1. Therefore we say,

CONTINUED Condition 3: As discussed before (see condition 2), the value of is 3 which we saw is equal to f (1) (see condition 1). Therefore, the function is continuous at x = 1 (though not differentiable).

§ 1.6 Some Rules for Differentiation

Section Outline Some Rules of Differentiation Differentiation
The Derivative as a Rate of Change

Rules of Differentiation

Differentiation EXAMPLE SOLUTION Differentiate
This is the given function. We begin to differentiate. Rewrite the rational expression with a negative exponent. Use the General Power Rule taking x3 + x + 1 to be g(x). Use the Sum Rule.

Differentiation CONTINUED Differentiate. Simplify. Simplify. Simplify.

Differentiation EXAMPLE SOLUTION Differentiate
This is the given function. We begin to differentiate. Rewrite the rational expression with a negative exponent. Use the Constant Multiple Rule. Use the General Power Rule taking to be g(x).

Differentiation CONTINUED Use the Sum Rule and rewrite as x1/2.
Differentiate. Simplify. Simplify. Simplify.

The Derivative as a Rate of Change

The Derivative as a Rate of Change
EXAMPLE Let S(x) represent the total sales (in thousands of dollars) for month x in the year 2005 at a certain department store. Represent each statement below by an equation involving S or (a) The sales at the end of January reached $120,560 and were rising at the rate of $1500 per month. (b) At the end of March, the sales for this month dropped to $80,000 and were falling by about $200 a day (Use 1 month = 30 days). SOLUTION Since the sales at the end of January (the first month, so x = 1) reached $120,560 and S(x) represents the amount of sales for a given month, we have: S(1) = 120,560. Further, since the rate of change of sales (rate of change means we will use the derivative of S(x)) for the month of January is a positive $1500 per month, we have: (b) At the end of March (the third month, so x = 3), the sales dropped to $80,000. Therefore, sales for the month of March was $80,000. That is: S(3) = 80,000. Additionally, since sales were dropping by $200 per day during March, this means that the rate of change of the function S(x) was (30 days) x (-200 dollars) = dollars per month. Therefore, we have:

§ 1.7 More About Derivatives

Section Outline Differentiating Various Independent Variables
Computing Second Derivatives Second Derivatives Evaluated at a Point Marginal Cost Marginal Revenue Marginal Profit

Differentiating Various Independent Variables
EXAMPLE Find the first derivative. SOLUTION We first note that the independent variable is t and the dependent variable is T. This is significant inasmuch as they are considered to be two totally different variables, just as x and y are different from each other. We now proceed to differentiate the function. This is the given function. We begin to differentiate. Use the Sum Rule. Use the General Power Rule.

Differentiating Various Independent Variables
CONTINUED Finish differentiating. Simplify. Simplify.

Second Derivatives EXAMPLE SOLUTION
Find the first and second derivatives. SOLUTION This is the given function. This is the first derivative. This is the second derivative.

Second Derivatives Evaluated at a Point
EXAMPLE Compute the following. SOLUTION Compute the first derivative. Compute the second derivative. Evaluate the second derivative at x = 2.

Marginal Cost

Marginal Cost EXAMPLE SOLUTION
Let C(x) be the cost (in dollars) of manufacturing x bicycles per day in a certain factory. Assume C(50) = 5000 and Estimate the cost of manufacturing 51 bicycles per day. SOLUTION We will first use the additional cost formula for manufacturing 1 more bicycle per day beyond the cost of producing 50 bicycles per day. We already know it costs $5000 to produce 50 bicycles per day since C(50) = So we wish to determine how much more, beyond that $5000, it costs to produce 51 bicycles. Therefore, we estimate the cost of manufacturing 51 bicycles to be $5045.

Marginal Revenue & Marginal Profit

Marginal Revenue EXAMPLE SOLUTION
Suppose the revenue from producing (and selling) x units of a product is given by R(x) = 3x – 0.01x2 dollars. (a) Find the marginal revenue at a production level of 20. (b) Find the production levels where the revenue is $200. SOLUTION (a) Since we are looking for the marginal revenue at a production level of 20, and we have an equation for R(x), we will simply find This is the given revenue function. This is the marginal revenue function. Evaluate the marginal revenue function at x = 20. Therefore, the marginal revenue at a production level of 20 is 2.6.

Marginal Revenue CONTINUED
(b) To find the production levels where the revenue is $200 we need to use the revenue function and replace revenue, R(x), with 200 and then solve for x. This is the given revenue function. Replace R(x) with 200. Get everything on the left side of the equation. Multiply everything by 100. Factor. Solve for x. Therefore, the production levels for which revenue is $200 are x = 100 and x = 200 units produced.

§ 1.8 The Derivative as a Rate of Change

Section Outline Average Rate of Change Instantaneous Rate of Change
Average Velocity Position, Velocity, and Acceleration Approximating the Change in a Function

Average Rate of Change

Average Rate of Change EXAMPLE SOLUTION
Suppose that f (x) = -6/x. What is the average rate of change of f (x) over the interval 1 to 1.2? SOLUTION The average rate of change over the interval is

Instantaneous Rate of Change

Instantaneous Rate of Change
EXAMPLE Suppose that f (x) = -6/x. What is the (instantaneous) rate of change of f (x) when x = 1? SOLUTION The rate of change of f (x) at x = 1 is equal to We have That is, the rate of change is 6 units per unit change in x.

Average & Instantaneous Rates of Change
EXAMPLE Refer to the figure below, where f (t) is the percentage yield (interest rate) on a 3-month T-bill (U.S. Treasury bill) t years after January 1, 1980. (a) What was the average rate of change in the yield from January 1, 1981 to January 1, 1985? (b) How fast was the percentage yield rising on January 1, 1989? (c) Was the percentage yield rising faster on January 1, 1980 or January 1, 1989?

CONTINUED SOLUTION (a) To determine the average rate of change in the yield from January 1, 1981 to January 1, 1985, we must first determine the coordinates of the two points that correspond to the two given dates. They are (1, 14) and (5, 7). Now we use the average rate of change formula. Therefore, the average rate of change in the yield from January 1, 1981 to January 1, 1985 is -7/4. (b) To determine how fast the percentage yield was rising on January 1, 1989, we must determine the instantaneous rate of change of f (t) when t = 9 (corresponding to January 1, 1989). This means that we must find the slope of the tangent line to the graph of f (t) where t = 9. The tangent line is on the graph and so we need only determine any two points on the tangent line. Using the coordinates of these two points, we will calculate the slope of the tangent line and that will be the instantaneous rate of change that we seek. Notice that two of the points on the tangent line are (5, 5) and (11, 10). Using these points we will now calculate the slope of the tangent line.

CONTINUED Therefore, the rate at which the percentage yield was rising on January 1, 1989 was 5/6. (c) To determine if the percentage yield was rising faster on January 1, 1980 or January 1, 1989, we would need to know the slopes of the tangent lines corresponding to t = 0 (January 1, 1980) and t = 9 (January 1, 1989). Although we already have this information for t = 9 (see part (b)), we do not yet have this information for t = 0. Therefore, we would first need to draw a tangent line to the graph corresponding to t = 0. This is done below.

CONTINUED Obviously, finding the coordinates of two points on this tangent line might prove a little difficult. However, notice that the slopes of the two tangent lines (all we’re really interested in are their slopes) are not remotely similar (that is, the tangent lines are not close to being parallel). Therefore, in this circumstance, it would be sufficiently appropriate to notice that the blue tangent line (corresponding to t = 0) has a steeper slope and therefore the rate of change was greater on January 1, 1980 than it was on January 1, 1989. NOTE: Use this technique of “eye-balling” a graph only when absolutely necessary and only with great care.

Average Velocity Definition Example
Average Velocity: Given a position function s(t), the average velocity from time t = a to t = a + h is Suppose a car is 3 miles from its starting point after 5 minutes and 7 miles from its starting point after an additional 6 minutes (after a total of 11 minutes). The average velocity of the car between the two given locations is miles per minute where a = 5 and h = 6.

Position, Velocity & Acceleration
s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.

EXAMPLE A toy rocket fired straight up into the air has height s(t) = 160t – 16t2 feet after t seconds. (a) What is the rocket’s initial velocity (when t = 0)? (b) What is the velocity after 2 seconds? (c) What is the acceleration when t = 3? (d) At what time will the rocket hit the ground? (e) At what velocity will the rocket be traveling just as it smashes into the ground? SOLUTION (a) To determine what the rocket’s initial velocity is, we must first have a velocity function and then evaluate t = 0 in that function. This is the given position function. Differentiate to get the velocity function.

CONTINUED Now replace t with 0 and evaluate. Therefore, the initial velocity of the rocket is 160 feet per second. (b) To determine the velocity after 2 seconds we evaluate v(2). This is the velocity function. Replace t with 2 and evaluate. Therefore, the velocity of the rocket after 2 seconds is 96 feet per second. (c) To determine the acceleration when t = 3, we must first find the acceleration function. This is the velocity function. Differentiate to get the acceleration function. Since the acceleration function is a constant function, the acceleration of the rocket is a constant ft/s2. Therefore, the acceleration when t = 3 is -32ft/s2.

CONTINUED (d) To determine at what time the rocket will hit the ground we must determine what we know about the position, velocity, or acceleration of the rocket when the rocket hits the ground. We know that at the time of impact, the position of the rocket will be 0 feet above the ground. Therefore, we will use the position function and replace s(t) with 0. This is the given position function. Replace s(t) with 0. Factor 16 out of both terms on the right and divide both sides by 16. Factor. Solve for t. Therefore, the rocket will be 0 feet above the ground at times t = 0 and t = 10. t = 0 corresponds to when the rocket first began its flight, so that would not be the solution. So, the rocket hit the ground after 10 seconds, when t – 10 = 0.

CONTINUED (e) To determine at what velocity the rocket will be traveling just as it smashes into the ground, we must use the velocity function. The question is, what do we use for t? From part (d), we know that the rocket will hit the ground at t = 10 seconds. Therefore, we will find v(10). This is the velocity function. Replace t with 10 and evaluate. Therefore, when the rocket hits the ground, it will be have a velocity of ft/s. That is, it will be traveling 160 ft/s in the downward direction.

Approximating the Change in a Function

EXAMPLE Suppose 5 mg of a drug is injected into the bloodstream. Let f (t) be the amount present in the bloodstream after t hours. Interpret f (3) = 2 and Estimate the number of milligrams of the drug in the bloodstream after 3½ hours. SOLUTION First, f (3) = 2 means that after 3 hours, 2 milligrams of the drug still remain in the bloodstream. Next, means that after 3 hours, the rate at which the amount of the drug is diminishing (because of the minus sign) within the bloodstream is ½ of a milligram per 1 hour. To estimate the number of milligrams of the drug in the bloodstream after 3½ hours, we will use the formula where a = 3 and h = ½ .

CONTINUED Therefore, approximate number of milligrams of the drug in the bloodstream after 3½ hours is 1.75.

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