   Chapter 5.3, Problem 47E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding an Equation of a Function In Exercises 47–50, find an equation of the function f that has the given derivative and whose graph passes through the given point. f ' ( x ) = 1 x 2 e 2 / x ;   ( 4 , 6 )

To determine

To calculate: The equation of function f(x) that has derivative f(x)=1x2e2/x and whose graph passes through the point (4,6).

Explanation

Given Information:

The first derivative f(x)=1x2e2/x.

The point is (4,6).

Formula used:

The general exponent rule of integrals:

eu(x)du(x)=ln|u(x)|+C

Here, u is function of x.

The property of Intro-differential:

df(x)dxdx=f(x)

Calculation:

Consider the derivative:

f(x)=1x2e2/x

Rewrite the integrand as:

df(x)dx=1x2e2/x

Apply, integration on both sides:

df(x)dxdx=1x2e2/xdx+Cf(x)=12e2x1

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