Question

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

Answer 1

Given:  

\[f(x) = x^2 - 4x + 7\]

Clearly,  

\[f(x)\]  being a polynomial function, is everywhere differentiable. The derivative of 

\[f\] at 

\[x\]  is given by:

\[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{\left( x + h \right)^2 - 4(x + h) + 7 - ( x^2 - 4x + 7)}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{x^2 + h^2 + 2xh - 4x - 4h + 7 - x^2 + 4x - 7}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h^2 + 2xh - 4h}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h(h + 2x - 4)}{h}\]
\[ \Rightarrow f'(x) = 2x - 4\]

Now, 

\[f'(5) = 2 \times 5 - 4 = 6\]
\[f'\left( \frac{7}{2} \right) = 2 \times \frac{7}{2} - 4 = 3\]

Therefore,   

\[f'(5) = 2 \times 3 = 2f'\left( \frac{7}{2} \right)\]

 Hence proved.