/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Let \(f(x)=2 e^{x}-6 x\). a. F... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 42

Let \(f(x)=2 e^{x}-6 x\). a. Find all points on the graph of \(f\) at which the tangent line is horizontal. b. Find all points on the graph of \(f\) at which the tangent line has slope 12.

Short Answer

Expert verified
Answer: a) There is a horizontal tangent line (slope 0) at the point \((\ln{3}, 2 e^{\ln{3}} - 6 \ln{3})\). b) There is a tangent line with a slope of 12 at the point \((\ln{9}, 2 e^{\ln{9}} - 6 \ln{9})\).

Step by step solution

01

Find the first derivative of the function f(x)

In order to find the slope of the tangent line to the function \(f(x)\), we need to find its first derivative, denoted as \(f'(x)\). Since \(f(x) = 2 e^x - 6x\), we have the following cases: 1. The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function, so if \(c\) is a constant and \(g(x)\) is a function, then \((c \cdot g(x))' = c \cdot g'(x)\). 2. The derivative of \(e^x\) is \(e^x\) (a very special property of the exponential function). 3. The derivative of \(x^n\) (where n is a constant) is \(n x^{n-1}\). Applying these rules, we find the first derivative of \(f(x)\): $$f'(x) = (2 e^x - 6x)' = 2 (e^x)' - 6(x)' = 2 e^x - 6.$$
02

Solve for horizontal tangent lines (slope = 0)

Now we will find all points where the tangent line to the graph of \(f(x)\) has a horizontal slope, which means that the slope of the tangent line is 0. So we need to solve the equation \(f'(x) = 0\): $$2 e^x - 6 = 0.$$ Add 6 to both sides and divide by 2: $$e^x = 3.$$ Now, take the natural logarithm of both sides: $$x = \ln(3).$$ The horizontal tangent line occurs at the point \((\ln{3}, f(\ln{3}))\), which is: $$(\ln{3}, 2 e^{\ln{3}} - 6 \ln{3}).$$
03

Solve for tangent lines with slope = 12

Next, we will find all points where the tangent line to the graph of \(f(x)\) has a slope of 12. So we need to solve the equation \(f'(x) = 12\): $$2 e^x - 6 = 12.$$ Add 6 to both sides and divide by 2: $$e^x = 9.$$ Now, take the natural logarithm of both sides: $$x = \ln(9).$$ The tangent line with a slope of 12 occurs at the point \((\ln(9), f(\ln(9)))\), which is: $$(\ln{9}, 2 e^{\ln{9}} - 6 \ln{9}).$$ In summary: a) The graph of \(f(x)\) has a horizontal tangent line at the point \((\ln{3}, 2 e^{\ln{3}} - 6 \ln{3})\). b) The graph of \(f(x)\) has a tangent line with a slope of 12 at the point \((\ln{9}, 2 e^{\ln{9}} - 6 \ln{9})\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Given the function \(f,\) find the slope of the line tangent to the graph of \(f^{-1}\) at the specified point on the graph of \(f^{-1}\) . $$f(x)=-x^{2}+8 ;(7,1)$$

Use any method to evaluate the derivative of the following functions. $$f(z)=z^{2}\left(e^{3 z}+4\right)-\frac{2 z}{z^{2}+1}$$

Prove that \(\frac{d^{2 n}}{d x^{2 n}}(\sin x)=(-1)^{n} \sin x\) and \(\frac{d^{2 n}}{d x^{2 n}}(\cos x)=(-1)^{n} \cos x\)

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=\frac{x}{x+5}$$

An observer stands \(20 \mathrm{m}\) from the bottom of a 10 -m-tall Ferris wheel on a line that is perpendicular to the face of the Ferris wheel. The wheel revolves at a rate of \(\pi\) rad/min and the observer's line of sight with a specific seat on the wheel makes an angle \(\theta\) with the ground (see figure). Forty seconds after that seat leaves the lowest point on the wheel, what is the rate of change of \(\theta ?\) Assume the observer's eyes are level with the bottom of the wheel.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.