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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.2.25b
Chapter 3, Problem 3.2.25b

21–30. Derivatives
b. Evaluate f'(a) for the given values of a.
f(x) = 1/x+1; a = -1/2;5

Verified step by step guidance
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Step 1: Identify the function f(x) = \(\frac{1}{x}\) + 1 and the points a = -\(\frac{1}{2}\) and a = 5 where we need to evaluate the derivative f'(a).
Step 2: Find the derivative of the function f(x). The function can be rewritten as f(x) = x^{-1} + 1. Use the power rule for derivatives, which states that \(\frac{d}{dx}\)[x^n] = nx^{n-1}, to find f'(x).
Step 3: Apply the power rule to f(x) = x^{-1} + 1. The derivative of x^{-1} is -x^{-2}, and the derivative of a constant (1) is 0. Therefore, f'(x) = -x^{-2}.
Step 4: Evaluate f'(x) at a = -\(\frac{1}{2}\). Substitute x = -\(\frac{1}{2}\) into f'(x) = -x^{-2} to find f'(-\(\frac{1}{2}\)).
Step 5: Evaluate f'(x) at a = 5. Substitute x = 5 into f'(x) = -x^{-2} to find f'(5).

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Key Concepts

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

Derivatives

A derivative represents the rate of change of a function with respect to its variable. It is a fundamental concept in calculus that provides information about the slope of the tangent line to the curve of the function at a given point. The derivative can be computed using various rules, such as the power rule, product rule, and quotient rule, depending on the form of the function.
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Derivatives

Quotient Rule

The quotient rule is a method for finding the derivative of a function that is the ratio of two other functions. If f(x) = g(x)/h(x), the derivative f'(x) is given by f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2. This rule is essential when dealing with functions that involve division, such as the function f(x) = 1/(x+1) in the given question.
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Evaluating Derivatives at Specific Points

Evaluating the derivative at a specific point involves substituting the value of that point into the derivative function. This process provides the slope of the tangent line to the function at that particular point, which can be useful for understanding the behavior of the function. In this case, we need to compute f'(-1/2) and f'(5) to analyze the function's behavior at these values.
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Related Practice
Textbook Question

Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

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b. h′(2)h^{\(\prime\)}\(\left\)(2\(\right\))

Textbook Question

Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = √3x; a= 12

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Textbook Question

Vertical tangent lines

b. Does the curve have any horizontal tangent lines? Explain.

Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

x = e^y; (2, ln 2)

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Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(s) = 4s³+3s; a= -3, -1

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Textbook Question

Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+32t+48.

b. When does the stone reach its highest point?

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