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Question
If f is defined by \[f\left( x \right) = x^2 - 4x + 7\] , show that \[f'\left( 5 \right) = 2f'\left( \frac{7}{2} \right)\]
Solution
Given:
Clearly,
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{\left( x + h \right)^2 - 4(x + h) + 7 - ( x^2 - 4x + 7)}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{x^2 + h^2 + 2xh - 4x - 4h + 7 - x^2 + 4x - 7}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h^2 + 2xh - 4h}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h(h + 2x - 4)}{h}\]
\[ \Rightarrow f'(x) = 2x - 4\]
Now,
\[f'(5) = 2 \times 5 - 4 = 6\]
\[f'\left( \frac{7}{2} \right) = 2 \times \frac{7}{2} - 4 = 3\]
Therefore,
Hence proved.
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