Question

Given $f\left(x\right)=4x^8,\mathrm{then}$

(a) $f\text{'}\left(\frac{1}{2}\right)=f\text{'}\left(-\frac{1}{2}\right)$

(b) $f\left(\frac{1}{2}\right)=-f\text{'}\left(-\frac{1}{2}\right)$

(c) $f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)$

(d) $f\left(\frac{1}{2}\right)=f\text{'}\left(-\frac{1}{2}\right)$

Answer 1

$$\begin{gathered} \left(\mathrm{c}\right)f\left(\frac{-1}{2}\right)=f\left(\frac{-1}{2}\right) \\ \mathrm{We}\mathrm{have},f\left(x\right)=4x^8 \\ \Rightarrow f\text{'}\left(x\right)=32x^7 \\ \mathrm{Now},f\left(\frac{1}{2}\right)=4{\left(\frac{1}{2}\right)}^8=4\left(\frac{1}{256}\right)=\frac{1}{64} \\ f\left(-\frac{1}{2}\right)=4{\left(-\frac{1}{2}\right)}^8=4\left(\frac{1}{256}\right)=\frac{1}{64} \\ f\text{'}\left(\frac{1}{2}\right)=32{\left(\frac{1}{2}\right)}^7=32\left(\frac{1}{128}\right)=\frac{1}{4} \\ f\text{'}\left(-\frac{1}{2}\right)=32{\left(-\frac{1}{2}\right)}^7=-32\left(\frac{1}{128}\right)=-\frac{1}{4} \end{gathered}$$