Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.10.9

Chapter 3, Problem 3.10.9

If f is a one-to-one function with f(3)=8 and f′(3)=7, find the equation of the line tangent to y=f^−1(x) at x=8.

Verified step by step guidance

First, understand that the problem involves finding the tangent line to the inverse function f^−1(x) at a specific point. We are given that f(3) = 8, which means that f^−1(8) = 3.

Recall that the derivative of the inverse function at a point x is given by the formula: (f^−1)'(x) = 1 / f'(f^−1(x)). This formula helps us find the slope of the tangent line to the inverse function.

Substitute the given values into the formula: Since f^−1(8) = 3 and f′(3) = 7, we have (f^−1)'(8) = 1 / 7. This is the slope of the tangent line at x = 8.

Now, use the point-slope form of the equation of a line, which is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line. Here, the point is (8, 3) and the slope m is 1/7.

Substitute the values into the point-slope form: y - 3 = (1/7)(x - 8). This equation represents the tangent line to y = f^−1(x) at x = 8.

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Key Concepts

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

Inverse Function

An inverse function reverses the effect of the original function. If f(x) takes an input x and produces an output y, then the inverse function f^−1(y) takes y back to x. For a function to have an inverse, it must be one-to-one, meaning each output is produced by exactly one input.

Derivative of Inverse Functions

The derivative of an inverse function can be found using the formula (f^−1)'(y) = 1 / f'(x), where y = f(x). This relationship is crucial for finding the slope of the tangent line to the inverse function at a given point, as it connects the rates of change of the original and inverse functions.

Tangent Line Equation

The equation of a tangent line at a point on a curve can be expressed as y - f(a) = f'(a)(x - a), where (a, f(a)) is the point of tangency. This equation uses the slope of the function at that point and the coordinates to describe the line that just touches the curve without crossing it.

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Related Practice

Textbook Question

13–40. Evaluate the derivative of the following functions.

f(x) = 1/tan^−1(x²+4)

Textbook Question

Vertical tangent lines If a function f is continuous at a and lim x→a| f′(x)|=∞, then the curve y=f(x) has a vertical tangent line at a, and the equation of the tangent line is x=a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 71–72) is used. Use this information to answer the following questions.

Graph the following curves and determine the location of any vertical tangent lines.

a. x²+y² = 9

Textbook Question

27–76. Calculate the derivative of the following functions.

$y = {(x^{2} + 2 x + 7)}^{8}$

Textbook Question

9–61. Evaluate and simplify y'.

y = (2x−3)x^3/2

Textbook Question

Find an equation of the line tangent to the curve y = sin x at x = 0.

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Textbook Question

Find d²/dx² (sin x + cos x).