Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.1.70b

Chapter 7, Problem 7.1.70b

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.

70. y= x³/(x²+1), -1 ≤ x ≤ 1, x_0=1/2

Verified step by step guidance

Start with the given function: \(y = \frac{x^{3}}{x^{2} + 1}\), where \(-1 \leq x \leq 1\).

To find the inverse function \(g(y)\), we need to solve the equation \(y = \frac{x^{3}}{x^{2} + 1}\) for \(x\) in terms of \(y\).

Multiply both sides of the equation by \((x^{2} + 1)\) to clear the denominator: \(y(x^{2} + 1) = x^{3}\).

Rewrite this as a polynomial equation in \(x\): \(x^{3} - y x^{2} - y = 0\).

Solve this cubic equation for \(x\) as a function of \(y\). The solution(s) will define the inverse function(s) \(g(y)\). Since the original function is restricted to \(-1 \leq x \leq 1\), select the root that lies within this domain to define the inverse.

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Key Concepts

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

Inverse Functions

An inverse function reverses the effect of the original function, swapping inputs and outputs. For a function f(x), its inverse g(y) satisfies g(f(x)) = x. Finding the inverse involves solving y = f(x) for x in terms of y, which may require algebraic manipulation and domain restrictions to ensure the inverse is well-defined.

Implicit Differentiation and Derivatives of Inverse Functions

When the inverse function is not given explicitly, implicit differentiation helps find its derivative. The derivative of the inverse at a point is the reciprocal of the derivative of the original function at the corresponding point, i.e., g'(y) = 1 / f'(x). This relationship is crucial for understanding tangent lines and rates of change of inverse functions.

Tangent Line Approximation

Tangent line approximation uses the derivative at a point to approximate the function near that point with a linear function. For a function f at x₀, the tangent line is y = f(x₀) + f'(x₀)(x - x₀). This concept extends to inverse functions, allowing approximation of g(y) near y₀ using the derivative of g.

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Related Practice

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b. natural logarithms.

71. ∫(from 1/5 to 3/13)dx/(x√(1-16x²))

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Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.

4. b. arcsin(-1/√2)

Textbook Question

75. b. Identify the function’s local and absolute extreme values, if any, saying where they occur.

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Find the volumes of the solids in Exercises 135 and 136.

135. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross-sections perpendicular to the x-axis are

b. vertical squares whose base edges run from the curve y=-1/√(1+x²) to the curve y=1/√(1+x²).

Textbook Question

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.

68. y= (3x+2)/(2x-11), -2 ≤ x ≤ 2, x_0=1/2