/*! 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} Q. 72 In Exercises 72–77, find a fun... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 72–77, find a function f that has the given derivative f. In each case you can find the answer with an educated guess-and-check process. (Some of these exercises involve hyperbolic functions.)

f'(x)=2x1+4x2

Short Answer

Expert verified

The function that has the derivative f'(x)=2x1+4x2is f(x)=121+4x2.

Step by step solution

01

Step 1. Given Information.

The derivative:

f'(x)=2x1+4x2

02

Step 2. Guess the formula for the given derivative.

We know that,

ddx(sinh-1x)=11+x2

So we can conclude that the derivative will be of the form of hyperbolic sinefunction.

03

Step 3. Check guess-and-check process.

f(x)=sinh-1(2x)f'(x)=11+x2.(2)=21+x2

which is not a given derivative.

04

Step 4. Check another function.

Consider the function,

f(x)=121+4x2f'(x)=ddx(12(1+4x2))12f'(x)=12.[12(1+4x2)-12.8x]=12[4x1+4x2]=2x1+4x2

which is the given derivative.

So the function is f(x)=121+4x2

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 91Ó°ÊÓ!

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

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.

The line tangent to the graph of y=4x+3 at the point(-2,-5)

A bowling ball dropped from a height of 400feet will be s(t)=400-16t2feet from the ground after tseconds. Use a sequence of average velocities to estimate the instantaneous velocities described below:

After t=1 seconds, with h=0.5,h=0.25,h=-0.5andh=-0.2

Velocity v(t) is the derivative of position s(t). It is also true that acceleration a(t) (the rate of change of velocity) is the derivative of velocity. If a race car’s position in miles t hours after the start of a race is given by the function s(t), what are the units of s(1.2)? What are the units and real-world interpretation of v(1.2)? What are the units and real-world interpretations of a(1.2)?

Every morning Linda takes a thirty-minute jog in Central Park. Suppose her distance s in feet from the oak tree on the north side of the park tminutes after she begins her jog is given by the function s(t)shown that follows at the left, and suppose she jogs on a straight path leading into the park from the oak tree.

(a) What was the average rate of change of Linda’s distance from the oak tree over the entire thirty-minute jog? What does this mean in real-world terms?

(b) On which ten-minute interval was the average rate of change of Linda’s distance from the oak tree the greatest: the first 10minutes, the second 10minutes, or the last10minutes?

(c) Use the graph of s(t)to estimate Linda’s average velocity during the 5-minute interval fromt=5tot=10. What does the sign of this average velocity tell you in real-world terms?

(d) Approximate the times at which Linda’s (instantaneous) velocity was equal to zero. What is the physical significance of these times?

(e) Approximate the time intervals during Linda’s jog that her (instantaneous) velocity was negative. What does a negative velocity mean in terms of this physical example?

Use the definition of the derivative to find ffor each function fin Exercises 39-54

f(x)=x3x+1

See all solutions

Recommended explanations on Math Textbooks

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.