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In the mean value theorem, $f(b)-f(a)=(b-a) f^{\prime}(c)$ if $a=4, b=9$ and $f(x)=\sqrt{x}$ then the value of $c$ is
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The correct answer is:
$6.25$
$f(x)=\sqrt{x}$
$\begin{aligned}& \therefore f(a)=\sqrt{4}=2, \quad f(b)=\sqrt{9}=3 ; \quad f^{\prime}(x)=\frac{1}{2 \sqrt{x}} \\
& \text { Also, } f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{3-2}{9-4}=\frac{1}{5}\end{aligned}$
Also, $\therefore \frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow c=\frac{25}{4}=6.25$
$\begin{aligned}& \therefore f(a)=\sqrt{4}=2, \quad f(b)=\sqrt{9}=3 ; \quad f^{\prime}(x)=\frac{1}{2 \sqrt{x}} \\
& \text { Also, } f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{3-2}{9-4}=\frac{1}{5}\end{aligned}$
Also, $\therefore \frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow c=\frac{25}{4}=6.25$
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