If $f$ is defined by $f(x)=x^{2}-4 x+7$, show that $f^{\prime}(5)=2 f^{\prime}\left(\frac{7}{2}\right)$
Given: $f(x)=x^{2}-4 x+7$
Clearly, $f(x)$ being a polynomial function, is everywhere differentiable. The derivative of $f$ at $x$ is given by:
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{2}-4(x+h)+7-\left(x^{2}-4 x+7\right)}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{2}+h^{2}+2 x h-4 x-4 h+7-x^{2}+4 x-7}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h^{2}+2 x h-4 h}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h(h+2 x-4)}{h}$
$\Rightarrow f^{\prime}(x)=2 x-4$
Now,
$f^{\prime}(5)=2 \times 5-4=6$
$f^{\prime}\left(\frac{7}{2}\right)=2 \times \frac{7}{2}-4=3$
Therefore, $f^{\prime}(5)=2 \times 3=2 f^{\prime}\left(\frac{7}{2}\right)$
Hence proved.
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