Ch. 7 - Transcendental Functions

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

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.1.70c

Chapter 7, Problem 7.1.70c

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

Verified step by step guidance

First, identify the function given: \(f(x) = \frac{x^3}{x^2 + 1}\) and the point at which we want the tangent line: \(x_0 = \frac{1}{2}\).

Calculate the value of the function at \(x_0\) to find the point on the curve: compute \(f\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^3}{\left(\frac{1}{2}\right)^2 + 1}\).

Find the derivative \(f'(x)\) using the quotient rule: if \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). Here, \(u(x) = x^3\) and \(v(x) = x^2 + 1\).

Compute \(u'(x) = 3x^2\) and \(v'(x) = 2x\), then substitute into the quotient rule formula to get \(f'(x)\).

Evaluate the derivative at \(x_0 = \frac{1}{2}\) to find the slope of the tangent line: \(m = f'\left(\frac{1}{2}\right)\). Then use the point-slope form of a line: \(y - f\left(\frac{1}{2}\right) = m \left(x - \frac{1}{2}\right)\) to write the equation of the tangent line.

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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 curve at that point. To find the tangent line equation, compute the derivative at x₀ to get the slope, then use the point-slope form y - f(x₀) = f'(x₀)(x - x₀). This line approximates the function near x₀.

Quotient Rule for Derivatives

When differentiating a function expressed as a quotient of two functions, use the quotient rule: (u/v)' = (u'v - uv') / v². Here, u = x³ and v = x² + 1, so find u' and v' separately, then apply the rule to find f'(x).

Evaluating Functions and Derivatives at a Point

After finding the derivative function, substitute the given x₀ value to find the slope at that point. Also, evaluate the original function at x₀ to get the point coordinates. These values are essential to write the tangent line equation accurately.

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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)

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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

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.

67. y= √(3x-2), 2/3 ≤ x ≤ 4, x_0=3

Textbook Question

3. Use the properties of logarithms to write the expressions in Exercises 3 and 4 as a single term.

b. ln(3x² - 9x) + ln(1/3x)

Textbook Question

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

Textbook Question

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