[tex]f(x)=x+\ln x[/tex] is differentiable for all [tex]x>0[/tex], so by the mean value theorem there is some [tex]c\in[1,4][/tex] for which
[tex]f'(c)=\dfrac{f(4)-f(1)}{4-1}[/tex]
We have derivative
[tex]f'(x)=1+\dfrac1x[/tex]
so that [tex]c[/tex] satisfies
[tex]1+\dfrac1c=\dfrac{4+\ln4-1}3\implies\boxed{c=\frac3{\ln4}}[/tex]
which can be rewritten using the change of base formula, if you wish:
[tex]\dfrac3{\ln 4}=\dfrac{\ln e^3}{\ln 4}=\log_4 e^3=3\log_4e[/tex]
The value will be "C = 2.164" for which the instantaneous rate of change is similar to the average rate.
Given:
Instantaneous rate of x=c is,
[tex]f'(x)_{x=c} = \frac{d}{dx} (x+ l_n x)[/tex] at x=c
[tex]=1+\frac{1}{x}[/tex] at x=c
[tex]=1+\frac{1}{c}[/tex] ...(2)
Now,
The average rate of change over [1, 4] will be:
→ [tex]\frac{f(4)-f(1)}{4-1}= \frac{(4+ l_n 4)-(1+l_n 1)}{3}[/tex]
[tex]= \frac{4+l_n4 -1}{3}[/tex]
[tex]= \frac{3+l_n 4}{3}[/tex] ...(3)
Since both are equal,
From equation "2" and "3", we get
→ [tex]\frac{3+ l_n 4}{3}=1+\frac{1}{C}[/tex]
→ [tex]1+\frac{l_n 4}{3} = 1+\frac{1}{C}[/tex]
→ [tex]\frac{l_n4}{3} = \frac{1}{c}[/tex]
→ [tex]C = \frac{3}{l_n 4}[/tex]
[tex]= \frac{3}{1.38629}[/tex]
[tex]= 2.164[/tex]
Thus the above answer is right.
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