Choose the correct alternative in the following:
Given $f(x)=4 x^{8}$, then
A. $f^{\prime}\left(\frac{1}{2}\right)=f^{\prime}\left(-\frac{1}{2}\right)$
B. $f\left(\frac{1}{2}\right)=f^{\prime}\left(-\frac{1}{2}\right)$
C. $f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)$
D. $f\left(\frac{1}{2}\right)=f^{\prime}\left(-\frac{1}{2}\right)$
$f(x)=4 x^{8}$
$f^{\prime}(x)=32 x^{7}$
Consider option (A)
$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=32\left(\frac{1}{2}\right)^{7}=32\left(\frac{1}{128}\right)=4$
$\mathrm{f}^{\prime}\left(-\frac{1}{2}\right)=32\left(-\frac{1}{2}\right)^{7}=32\left(-\frac{1}{128}\right)=-4$
$\mathrm{f}^{\prime}\left(\frac{1}{2}\right) \neq \mathrm{f}^{\prime}\left(-\frac{1}{2}\right)$
Consider option (B)
$f\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{8}=4\left(\frac{1}{256}\right)=64$
$f^{\prime}\left(-\frac{1}{2}\right)=32\left(-\frac{1}{2}\right)^{7}=32\left(-\frac{1}{128}\right)=-4$
$f\left(\frac{1}{2}\right) \neq f^{\prime}\left(-\frac{1}{2}\right)$
Consider option (C)
$f\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^{8}=4\left(\frac{1}{256}\right)=64$
$f\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^{8}=4\left(\frac{1}{256}\right)=64$
$\therefore f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)=(C)$
Consider option (D)
$f\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{8}=4\left(\frac{1}{256}\right)=64$
$f^{\prime}\left(-\frac{1}{2}\right)=32\left(-\frac{1}{2}\right)^{7}=32\left(-\frac{1}{128}\right)=-4$
$f\left(\frac{1}{2}\right) \neq f^{\prime}\left(-\frac{1}{2}\right)$