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Verify Mean Value Theorem, if $f(x)=x^2-4 x-3$ in the interval $[a, b]$, where $a=1$ and $b=4$.
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$f(x)=x^2-4 x-3$. It being a polynomial it is continuous in the interval $[1,4]$ and derivable in $(1,4)$, So all the condition of mean value theorem hold.
then $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-4, \therefore \mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{c}-4$
$f(4)=16-16-3=-3, f(1)=1-4-3=-6$
Then there exist a value $\mathrm{c}$ such that
$\begin{aligned}
&f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow 2 c-4=\frac{-3-(6)}{4-1}=\frac{3}{3}=1 \\
&\therefore 2 c=4+1 \quad \text { or } \quad c=\frac{5}{2} \in(1,4)
\end{aligned}$
then $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-4, \therefore \mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{c}-4$
$f(4)=16-16-3=-3, f(1)=1-4-3=-6$
Then there exist a value $\mathrm{c}$ such that
$\begin{aligned}
&f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow 2 c-4=\frac{-3-(6)}{4-1}=\frac{3}{3}=1 \\
&\therefore 2 c=4+1 \quad \text { or } \quad c=\frac{5}{2} \in(1,4)
\end{aligned}$
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