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Chapter 5: Q.59 (page 442)

The function \(f(x)=\frac{2x^3+x}{(x^2+1)^2}\).

(a) Find the average value of \(f(x)\) on \([0,2]\).

(b) Graphically approximate a value of x for which f (x) is equal to its average value on [0, 2], and use this value to verify that your answer from part (a) is reasonable.

Short Answer

Expert verified

(a) \(0.6047\)

(b) Verified

Step by step solution

01

Part (a) Step 1: Given Information

Consider the function \(f(x)=\frac{2x^3+x}{(x^2+1)^2}\).

The average of the function \(f(x)\) on \([0,2]\).

\(f_{avg}=\frac{1}{b-a}\int_a^bf(x)dx\)

02

Part (b) Step 2: Find the average of a function

\(\begin{align*}f_{avg}&=\frac{1}{2-0}\int_0^2\frac{2x^3+x}{(x^2+1)^2}dx\\&=0.6047\end{align*}\)

03

Part (b) Step 1: Given Information

Consider the function \(f(x)=\frac{2x^3+x}{(x^2+1)^2}\).

The average value is \(0.6047\).

04

Part (b) Step 2: Find \((x)\)

\(\frac{2x^3+x}{(x^2+1)^2}=0.6047\)

Solve for \(x\). Using graphing calculator the obtained values are:

\(x=0.7\) and \(2.7\).

If check the difference of the \(x\) value will get the same as the interval \([0,2]\) difference.

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