Answer:
c = 2.35
Explanation:
The Mean Value Theorem is defined as:
[tex]\begin{gathered} f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \\ \text{and} \\ f(c)=\frac{1}{b-a}\int ^4_1f(x)dx \end{gathered}[/tex]
With the given f(x), we have:
[tex]\begin{gathered} f(c)=\frac{1}{b-a}\int ^4_1\frac{x}{x+2}dx \\ \\ f(c)=\frac{1}{4-1}\int ^4_1(1-\frac{2}{x+2})dx \\ \\ =\frac{1}{3}\lbrack x-2\ln (x+2)\rbrack\begin{cases}4 \\ 1\end{cases} \\ \\ =\frac{1}{3}\lbrack(4-2\ln (4+2))-(1-2\ln (1+2))\rbrack \\ \\ =1-\frac{2}{3}\ln (2) \\ \\ =0.54 \end{gathered}[/tex]
We have
[tex]\begin{gathered} f(c)=0.54 \\ \frac{c}{c+2}=0.54 \\ \\ c=0.54(c+2) \\ c-0.54c=2(0.54) \\ 0.46c=1.08 \\ c=\frac{1.08}{0.46}=2.35 \end{gathered}[/tex]
