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By Michele Bunch, Kimberly Duane and Mark Duane. Extreme Values Critical values – any value along f(x) where f’(x) = 0 OR f’(x) is undefined These help.

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Presentation on theme: "By Michele Bunch, Kimberly Duane and Mark Duane. Extreme Values Critical values – any value along f(x) where f’(x) = 0 OR f’(x) is undefined These help."— Presentation transcript:

1 By Michele Bunch, Kimberly Duane and Mark Duane

2 Extreme Values Critical values – any value along f(x) where f’(x) = 0 OR f’(x) is undefined These help you determine extreme values: Relative max/min → over ENTIRE FUNCTION Note: NOT ALL CV’S ARE RELATIVE EXTREMA! Absolute max/min → over an INTERVAL If f(x) is continuous on a closed interval, extrema could be at CV’s OR at endpoints – check them all! Key words If it asks WHERE the max/min is, it wants the x value If it asks WHAT it is, then it wants the value

3 Find the absolute max and min on the interval [-3,2] Critical Values Endpoints Because x=4 is not in the interval, you don’t have to test it IN THIS CASE!! Absolute max: x=2 Absolute min: x=-3

4 1 st Derivative Test This is a test that uses the first derivative to help you find relative maxes and mins. 1. Find critical values and create a number line of f’(x) with these values. 2. Use f’(x) to determine if f’(x) is positive or negative. -- If f’(x) is positive, f(x) is increasing -- If f’(x) is negative, f(x) is decreasing 3. If f’(x) changes from positive to negative, the CV is a local max. If f’(x) changes from negative to positive, the CV is a local min. 14 x = 1 is a local max; x = 4 is a local min

5 2 nd Derivative Test Still a test for determining local max and min! 1. Take the 2 nd derivative, set it equal to zero, and solve. 2. Place these values on a number line for f”(x) and use f”(x) to determine if f”(x) is positive or negative. 5/2 3. For the critical values: -- If f”(CV) is positive, then the CV is a local min -- If f”(CV) is negative, then the CV is a local max

6 Concavity Concave UpConcave Down f’(x) is increasing f”(x) is positive This means the slopes are always increasing f’(x) is decreasing f”(x) is negative This means the slopes are always decreasing Looks like this:

7 f(x), f’(x), and f”(x) The graph of the derivative is essentially the slope of the tangent line as it travels along the curve. For every point on f(x) that has a horizontal tangent line, f’(x) will cross the x axis at that same point. If the slope of the tangent line is positive and increasing for f(x), f’(x) will be positive (and vice versa) f’(x) and f”(x) have the same relationship as f(x) and f”(x)

8 Mean Value Theorem Given a differentiable function over an interval, there is at least one point on the curve where the derivative is equal to the average derivative of the entire interval

9 Examples Find all values of “c” that satisfy the MVT. 1) [-7,1] f(-7)=-2 f(1)=0 f’(c)=2/8=1/4 2) y=x ⁻ ¹ [1,4] f(1)=1 f(4)=1/2 f’(c)=-1/6

10 Optimization Using related formulas to find the maximum or minimum solution to a problem Steps: 1) Sketch the problem 2) Write equations (target and constraints) 3) Combine equations 4) Determine domain 5) Find 1 st derivative and critical points 6) Find maximum/minimum 7) Give answer in a sentence

11 Example One Two pens are to be constructed using a total of 900 feet of fencing. One pen is to be a square X by X and the other is to be a rectangle with one side twice as long as the other (X by 2X). Determine the dimensions of the pens so that the enclosed areas are as large as possible. x x 2x x So the dimensions with the maximum area are 105.9 ft by 105.9 ft and 79.4 ft by 158.8 ft.

12 Example Two Given the curve y=x² in the first quadrant and a vertical line x=3, determine the inscribed rectangle of maximum area which has a right side on line x=3. (x,y) 3-x x²x² The rectangle is 2 by 4, so the area of the rectangle is 8.

13 Related Rates Finding the rate at which a quantity changes by comparing to a known rate of change Usually used in relation to time Steps: 1) Draw a picture 2) State wanted and known rates 3) Write formula to compare rates 4) Take the derivative 5) Plug in the known rate 6) Solve for unknown rate

14 Example One A 20 foot ladder is leaning against a building. The ladder is sliding down the wall at a constant rate of 2 ft/sec. At what rate is the angle between the ladder and the ground changing when the top of the ladder is 12 feet from the ground? The angle is changing at a rate of -1/8 radians per second. ɵ x h20

15 Example Two Water is flowing into and inverted cone at the rate of 5 in³/sec. If the cone has an altitude of 4 inches and a base radius of 3 inches, how fast is the water level rising when the water is 2 inches deep? The water level is rising at 20/(9π) inches per second. 4 h 3

16 Particle Motion Particle motion is described by a function and its derivatives. When given a function that describes a particles linear motion, the first derivative describes the particle's velocity and the absolute value of the first derivative gives speed. The second derivative gives the particles acceleration.

17 Example One Given the following equation that gives the position of a particle, find A) the velocity equation, B) the velocity at time 15 minutes, and C) the speed at 15 minutes. A) B)f’(15)=-8.5 m/s C)8.5 m/s

18 Example Two The velocity (in centimeters per second) of a blood molecule flowing through a capillary of radius 0.008 cm is given by the equation below. Find the acceleration when r = 0.004 The acceleration at r=0.004 is - 0.00008 centimeters per second squared.

19 Linear Approximation Linear approximation is using the slope of a tangent line to approximate a value that is close to the point of tangency. Δ f(x)=f’(a)-x (yes, the delta is upside down). Δf(x) = f(a + Δx) – f(a).

20 Example One The cube root of 27 is 3. How much larger is the cube root of 27.2? The derivative of f(x) is Δf '(x)= (1/27)(0.2) Δf '(x)= 1/135 So the cube root of 27.2 is 1/135 more than the cube root of 27 Estimated value =3.00740 Actual value=3.00739

21 Example Two The price of a bus pass between Albuquerque and Los Alamos is set at x dollars, a bus company takes in a monthly revenue of in thousands of dollars. Estimate the change in revenue from $50 to $53. ∆R'(50)=.5.5*∆x =.5*3 = 1.5 So by increasing the tickets by three dollars the company's revenue goes up by $1500 to 51500. The actual value for $53 is $514100.

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