Question

If f is defined by $f\left(x\right)=x^2-4x+7$, show that $f\text{'}\left(5\right)=2f\text{'}\left(\frac{7}{2}\right)$

Answer 1

Given: $f\left(x\right)=x^2-4x+7$

Clearly, $f\left(x\right)$ being a polynomial function, is everywhere differentiable. The derivative of $f$ at $x$ is given by:

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^2-4(x+h)+7-(x^2-4x+7)}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{x^2+h^2+2xh-4x-4h+7-x^2+4x-7}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{h^2+2xh-4h}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{h(h+2x-4)}{h} \\ \Rightarrow f\text{'}\left(x\right)=2x-4 \end{gathered}$$

Now,

$$\begin{gathered} f\text{'}\left(5\right)=2\times 5-4=6 \\ f\text{'}\left(\frac{7}{2}\right)=2\times \frac{7}{2}-4=3 \end{gathered}$$
Therefore, $f\text{'}\left(5\right)=2\times 3=2f\text{'}\left(\frac{7}{2}\right)$
Hence proved.