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**Chapter 3 Application of Derivatives**

3.1 Extreme Values of Functions Absolute maxima or minima are also referred to as global maxima or minima.

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Examples

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Extreme Value Theorem The requirements in Theorems 1 that the interval be closed and finite, and That the function be continuous, are key ingredients.

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**Local (Relative) Extreme Values**

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Finding Extrema

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Critical Point Theorem 2 says that a function’s first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. Hence the only places where a function f can possibly have an extreme value (local or global) are Interior points where f’=0 Interior points where f’ is undefined, Endpoints of the domain of f.

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**How to Find the Absolute Extrema**

Thus the only domain points where a function can assume extreme values are critical points and endpoints.

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Example Example. Find the absolute maximum and minimum values of f(x)=10x(2-lnx) on the interval [1, ex]. Solution. We evaluate the function at the critical points and the endpoints and take the largest and the smallest of the resulting values. The first derivative is f’(x)=10(2-lnx)-10x(1/x)=10(1-lnx). Let f’(x)=0, we have x=e. Then Critical point value: f(e)=10e Endpoint values: f(1)=20, and f(e2)=0. So the function’s absolute maximum value is10e at x=e. The absolute minimum value is 0 and occurs at the right endpoint x=e2.

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3.2 The Mean Value Theorem

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**Mathematical Consequences**

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**3.3 Monotonic Functions and the First Derivative Test**

A function that is increasing or decreasing on an interval is said to be monotonic on the interval.

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**Graphs of functions Each tangent line Has positive slope.**

Has negative slope. Each tangent line Has zero slope.

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Theorem

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Example Example: Find the critical points of f(x)=x3 -12x-5 and identify the intervals on which f is increasing and on which f is decreasing. Solution: The function f is everywhere continuous and differentiable. The first derivative f ’(x)=3x2-12=3(x2-4)=3(x+2)(x-2) is zero at x=-2 and x=2. These critical points subdivide the domain of f into intervals (-, -2), (-2, 2) and (2, ) on which f ‘ is either positive or negative. We determine the sign Of f ‘ by evaluating f at a convenient point in each subinterval. Interval <x< <x< <x< f ‘ evaluated f ’(-3) = f ‘(0)= f ‘ (3) =15 Sign of f ‘ Behavior of f increasing decreasing increasing

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**First Derivative Test for Local Extrema**

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Example Find the critical points of f(x)=(x2-3)ex. Identify the intervals on which f is increasing and decreasing. Find the function’s local and absolute extreme values. Solution. f ’(x)=(x2+2x-3)ex. Since ex is never zero, f ‘(x) is zero iff x2+2x-3=0. That is, x=-3 and x=1. We can see that there is a local maximum about at x=-3 and a local minimum about at x=1. The local minimum value is also an absolute minimum, but there is no absolute maximum. The function increases on (-, -3) and (1, ) and deceases on (-3, 1).

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**3.4 Concavity and Curve Sketching**

Two ways to characterize the concavity of a differentiable function f on an open interval: f is concave up on an open interval if its tangent lines have increasing slopes on that interval and is concave down if they have decreasing slopes. f is concave up on an open interval if its graph lies above its tangent lines on that interval and is concave down if it lies below its tangent lines

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**Concavity and the Second Derivative Test for Concavity**

If y=f(x) is twice-differentiable, we will use f’’ and y’’ interchangeable When denoting the second derivative.

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Example Example: Find the intervals on which is concave up and the intervals on which it is concave down. Solution: Thus f(x) is concave up on the interval , and concave down on the interval

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Inflection Points

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**Second Derivative Test for Local Extrema**

This test requires us to know f’’ only at c itself and not in an interval about c. This makes the test easy to apply. However, this test is inconclusive if f’’=0 or if ‘’ does not exist at x=c. When this happens, use the First derivative Test for local extreme values.

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**Strategy for Graphing y=f(x)**

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**Graphical Behavior of Functions from Derivatives**

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**3.5 Parametrizations of Plane Curves**

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**Parametric Formula for dy/dx**

Example. Find dy/dx as a function of t if x = t - t2, y = t - t3. Solution.

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**3.7 Indeterminate Forms and L’Hopital’s Rule**

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**How to Use L’Hopital’ Rule**

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Example Example: Find Solution: The limit is a indeterminate form of type 0/0. Applying L’Hopital’s rule yields

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Example Example: Find Solution: The limit is a indeterminate form of type 0/0. Applying L’Hopital’s rule yields

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**Indeterminate Forms of Type /**

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Example Example: Find Solution: The limit is a indeterminate form of type Applying L’Hopital’s rule yields In fact, we can use LHopital’s rule to show that

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Example Example: Find Solution: The limit is a indeterminate form of type Applying L’Hopital’s rule yields Similar methods can be used to find the limit of f(x)/g(x) is an Indeterminate form of the types:

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3.8 Newton’s Method Newton’s method is a technique to approximate the solution to an equation f(x)=0. Essentially it uses tangent lines in place of the graph of y=f(x) near the points where f is zero. (A value of x where f is zero is a root of the function f and a solution of the equation f(x)=0.)

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**To find a root r of the equation f(x)=0,**

select an initial approximation x1. If f(x1)=0, then r=x1. Otherwise, use the root of the tangent line to the graph of f at x1 to approximate r. Call this intercept x2 . We can now treat x2 in the same way we did x1. If f(x2 )=0, then r= x2 . Otherwise, we construct the tangent line to the graph of f at x2, and take x3 to be the x-intercept of the tangent line. Continuing in this way, we can generate a succession of values x1,x2,, x3,,,x4…that will usually approach r.

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Example Use Newton’s Method to approximate the real solutions of x3-x-1=0 Solution: Let f(x)=x3-x-1. Since f(1)=-1 and f(2)=5, weh know by the Intermediate Value Theorem, there is a root in the interval (1, 2). We apply Newton’s method to f with the starting values x0 =1. The result are displayed in the following table and figure.

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**At n=5, we come to the result x6=x5=1. 324717957**

At n=5, we come to the result x6=x5= When xn+1=xn, Equation (1) shows that f(xn)=0. We have found a folution of f(x)=0 to nine decimals.

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3.9 Hyperbolic Functions

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**Identities and Derivatives for Hyperbolic Functions**

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**Inverses Hyperbolic Sine, Cosine, and Secant of x**

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**Inverse Hyperbolic Tangent, Cotangent, and Cosecant of x**

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**Identities and Derivatives of Inverse Hyperbolic Functions**

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