Question

If $f(x) = 2x^8$, then the correct statement is :

• A. $f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right)$
• B. $f\left(\frac{1}{2}\right) = -f\left(-\frac{1}{2}\right)$
• C. $f'\left(\frac{1}{2}\right) = -f'\left(-\frac{1}{2}\right)$
• D. $f\left(\frac{1}{2}\right) = -f\left(-\frac{1}{2}\right)$

Hint:

Even functions satisfy $f(x) = f(-x)$, which is true for the given function.

Answer 1

The Correct Option is A

The function $f(x) = 2x^8$ is an even function, confirmed by $f(x) = f(-x)$. Evaluating at $x = \frac{1}{2}$, we find: \[ f\left(\frac{1}{2}\right) = 2 \left(\frac{1}{2}\right)^8 = 2 \times \frac{1}{256} = \frac{1}{128} \] Likewise, \[ f\left(-\frac{1}{2}\right) = 2 \left(-\frac{1}{2}\right)^8 = \frac{1}{128} \] This demonstrates that $f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right)$.