If f is defined by

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.