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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.1.83
Chapter 4, Problem 4.1.83

{Use of Tech} Critical points and extreme values


a. Find the critical points of the following functions on the given interval. Use a root finder, if necessary.
b. Use a graphing utility to determine whether the critical points correspond to local maxima, local minima, or neither.
c. Find the absolute maximum and minimum values on the given interval, if they exist


h(x) (5-x)/(x² + 2x - 3) on [-10,10]

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1
To find the critical points of the function \( h(x) = \frac{5-x}{x^2 + 2x - 3} \), first compute the derivative \( h'(x) \). Use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = 5-x \) and \( v = x^2 + 2x - 3 \).
Calculate the derivatives: \( u' = -1 \) and \( v' = 2x + 2 \). Substitute these into the quotient rule to find \( h'(x) = \frac{(-1)(x^2 + 2x - 3) - (5-x)(2x + 2)}{(x^2 + 2x - 3)^2} \). Simplify the expression to find \( h'(x) \).
Set \( h'(x) = 0 \) to find the critical points. This involves solving the equation \( (-1)(x^2 + 2x - 3) - (5-x)(2x + 2) = 0 \). Use algebraic manipulation or a root finder to solve for \( x \).
Use a graphing utility to plot \( h(x) \) and visually inspect the critical points. Determine whether each critical point is a local maximum, local minimum, or neither by observing the behavior of the graph around these points.
To find the absolute maximum and minimum values on the interval \([-10, 10]\), evaluate \( h(x) \) at the critical points and at the endpoints \( x = -10 \) and \( x = 10 \). Compare these values to determine the absolute extrema.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are essential for identifying local maxima and minima, as they represent potential locations where the function's behavior changes. To find critical points, one typically differentiates the function and solves for the values of x that satisfy the derivative equation.
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Local Extrema

Local extrema refer to the highest or lowest points in a specific interval of a function. A local maximum is a point where the function value is greater than its neighboring values, while a local minimum is where it is lower. Determining local extrema often involves analyzing critical points and using the first or second derivative tests to classify them.
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Absolute Extrema

Absolute extrema are the overall highest or lowest values of a function on a given interval. To find these values, one must evaluate the function at its critical points and also at the endpoints of the interval. The largest value among these evaluations is the absolute maximum, while the smallest is the absolute minimum, providing a complete picture of the function's behavior over the specified range.
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