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 3.10.5

Chapter 3, Problem 3.10.5

Suppose f is a one-to-one function with f(2)=8 and f′(2)=4. What is the value of (f^−1)′(8)?

Verified step by step guidance

Step 1: Recall the formula for the derivative of the inverse function. If f is a one-to-one function and differentiable at a point, then the derivative of its inverse function at a point is given by: \((f^{-1})'(b) = \frac{1}{f'(a)}\), where \(f(a) = b\).

Step 2: Identify the given values in the problem. We know that \(f(2) = 8\) and \(f'(2) = 4\).

Step 3: Match the given values to the formula. Here, \(a = 2\) and \(b = 8\), so we need to find \((f^{-1})'(8)\).

Step 4: Substitute the known values into the formula. Using \((f^{-1})'(b) = \frac{1}{f'(a)}\), substitute \(f'(2) = 4\) into the formula to find \((f^{-1})'(8)\).

Step 5: Calculate the expression \((f^{-1})'(8) = \frac{1}{4}\). This gives the value of the derivative of the inverse function at the point where \(f(x) = 8\).

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Key Concepts

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

Inverse Function Theorem

The Inverse Function Theorem states that if a function f is continuously differentiable and has a non-zero derivative at a point, then its inverse function f⁻¹ is also differentiable at the corresponding point. Specifically, if f'(a) ≠ 0, then (f⁻¹)'(f(a)) = 1 / f'(a). This theorem is crucial for finding the derivative of an inverse function.

One-to-One Function

A one-to-one function, or injective function, is a function where each output is produced by exactly one input. This property ensures that the function has an inverse, as it guarantees that for every y in the range, there is a unique x in the domain such that f(x) = y. In this problem, knowing that f is one-to-one allows us to confidently apply the Inverse Function Theorem.

Derivative of a Function

The derivative of a function at a point measures the rate at which the function's value changes as its input changes. It is defined as the limit of the average rate of change as the interval approaches zero. In this context, f′(2) = 4 indicates that at x = 2, the function f is increasing at a rate of 4 units of output for every 1 unit of input, which is essential for calculating the derivative of the inverse function.

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Related Practice

Textbook Question

Watching an elevator An observer is 20 m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 20 m horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer’s line of sight makes with the horizontal (it may be positive or negative). Assuming the elevator rises at a rate of 5 m/s, what is the rate of change of the angle of elevation when the elevator is 10 m above the ground? When the elevator is 40 m above the ground? <IMAGE>

Textbook Question

Find the following higher-order derivatives.

d²/dx² (In(x² + 1))

Textbook Question

Find (f^−1)′(3), where f(x)=x³+x+1.

Textbook Question

Use implicit differentiation to find dy/dx.

sin xy = x+y

Textbook Question

Calculate the derivative of the following functions (i) using the fact that b^x = e^{xIn b} and (ii) using logarithmic differentiation. Verify that both answers are the same.

y = (4x+1)^{In x}

Textbook Question

90–93. {Use of Tech} Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph.

x(1−y²)+y³=0