Choose the correct alternative in the following:

Solution:

$f(x)=4 x^{8}$

$f^{\prime}(x)=32 x^{7}$

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)

$f\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{8}=4\left(\frac{1}{256}\right)=64$

$f^{\prime}\left(-\frac{1}{2}\right)=32\left(-\frac{1}{2}\right)^{7}=32\left(-\frac{1}{128}\right)=-4$

$f\left(\frac{1}{2}\right) \neq f^{\prime}\left(-\frac{1}{2}\right)$