Ch. 7 - Transcendental Functions

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

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.1.72c

Chapter 7, Problem 7.1.72c

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:

c. Find the equation for the tangent line to f at the specified point (x_0, f(x_0)).

72. y= 2-x-x³, -2 ≤ x ≤ 2, x_0 = 3/2

Verified step by step guidance

Identify the function given: \(y = 2 - x - x^{3}\) and the point at which to find the tangent line, \(x_0 = \frac{3}{2}\).

Calculate the value of the function at \(x_0\) to find the point of tangency: compute \(f\left(\frac{3}{2}\right) = 2 - \frac{3}{2} - \left(\frac{3}{2}\right)^{3}\).

Find the derivative of the function \(f(x)\) to get the slope of the tangent line: \(f'(x) = \frac{d}{dx} \left(2 - x - x^{3}\right)\).

Evaluate the derivative at \(x_0\) to find the slope of the tangent line at that point: calculate \(f'\left(\frac{3}{2}\right)\).

Use the point-slope form of the line equation with the point \(\left(x_0, f(x_0)\right)\) and slope \(f'(x_0)\): write the tangent line as \(y - f\left(\frac{3}{2}\right) = f'\left(\frac{3}{2}\right) \left(x - \frac{3}{2}\right)\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Key Concepts

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

Derivative and Tangent Line

The derivative of a function at a point gives the slope of the tangent line to the function's graph at that point. To find the tangent line equation, compute the derivative at x₀ to get the slope, then use the point-slope form with (x₀, f(x₀)). This line approximates the function near x₀.

Evaluating Functions and Derivatives at a Point

Evaluating the function and its derivative at a specific point x₀ involves substituting x₀ into the function and its derivative expression. This provides the coordinates of the point on the curve and the slope of the tangent line, both essential for writing the tangent line equation.

Using Computer Algebra Systems (CAS)

A CAS can symbolically compute derivatives and evaluate expressions, simplifying the process of finding tangent lines. It helps avoid manual calculation errors and speeds up solving, especially for complex functions like polynomials or when working with inverse functions.

Recommended video:

01:56

Finding Limits by Direct Substitution Example 1

Related Practice

Textbook Question

Evaluate the integrals in Exercises 67–74 in terms of

b. natural logarithms.

73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)

views

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:

c. Find the equation for the tangent line to f at the specified point (x_0, f(x_0)).

70. y= x³/(x²+1), -1 ≤ x ≤ 1, x_0=1/2

Textbook Question

Suppose that the function f and its derivative with respect to x have the following values at x=0, 1, 2, 3, and 4.

Assuming the inverse function f^(-1) is differentiable, find the slope of f^(-1)(x) at

c. x=3

Textbook Question

3. Which of the following functions grow faster than x² as x→∞? Which grow at the same rate as x²? Which grow slower?

c. √(x^4 + x^3)

Textbook Question

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

67. y= √(3x-2), 2/3 ≤ x ≤ 4, x_0=3

Textbook Question

c. Find the slopes of the tangent lines to the graphs of h and k at (2, 2) and (−2, −2).