Question

Given  \[f\left( x \right) = 4 x^8 , \text { then }\] _________________ .

पर्याय

• \[f'\left( \frac{1}{2} \right) = f'\left( - \frac{1}{2} \right)\]

• \[f\left( \frac{1}{2} \right) = - f'\left( - \frac{1}{2} \right)\]

• \[f\left( - \frac{1}{2} \right) = f\left( - \frac{1}{2} \right)\]

• \[f\left( \frac{1}{2} \right) = f'\left( - \frac{1}{2} \right)\]

Answer 1

\[\ f\left( \frac{- 1}{2} \right) = f\left( \frac{- 1}{2} \right)\]
\[\text{ We have }, f\left( x \right) = 4 x^8 \]
\[ \Rightarrow f'\left( x \right) = 32 x^7 \]
\[\text{ Now,} f\left( \frac{1}{2} \right) = 4 \left( \frac{1}{2} \right)^8 = 4\left( \frac{1}{256} \right) = \frac{1}{64}\]
\[ f\left( - \frac{1}{2} \right) = 4 \left( - \frac{1}{2} \right)^8 = 4\left( \frac{1}{256} \right) = \frac{1}{64}\]
\[f'\left( \frac{1}{2} \right) = 32 \left( \frac{1}{2} \right)^7 = 32\left( \frac{1}{128} \right) = \frac{1}{4}\]
\[f'\left( - \frac{1}{2} \right) = 32 \left( - \frac{1}{2} \right)^7 = - 32\left( \frac{1}{128} \right) = - \frac{1}{4}\]