Question

# Given $f\left(x\right)=4{x}^{8},\mathrm{then}$ (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) $f\left(\frac{1}{2}\right)=f\text{'}\left(-\frac{1}{2}\right)$

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Solution

## $\left(\mathrm{c}\right)f\left(\frac{-1}{2}\right)=f\left(\frac{-1}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have},f\left(x\right)=4{x}^{8}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=32{x}^{7}\phantom{\rule{0ex}{0ex}}\mathrm{Now},f\left(\frac{1}{2}\right)=4{\left(\frac{1}{2}\right)}^{8}=4\left(\frac{1}{256}\right)=\frac{1}{64}\phantom{\rule{0ex}{0ex}}f\left(-\frac{1}{2}\right)=4{\left(-\frac{1}{2}\right)}^{8}=4\left(\frac{1}{256}\right)=\frac{1}{64}\phantom{\rule{0ex}{0ex}}f\text{'}\left(\frac{1}{2}\right)=32{\left(\frac{1}{2}\right)}^{7}=32\left(\frac{1}{128}\right)=\frac{1}{4}\phantom{\rule{0ex}{0ex}}f\text{'}\left(-\frac{1}{2}\right)=32{\left(-\frac{1}{2}\right)}^{7}=-32\left(\frac{1}{128}\right)=-\frac{1}{4}\phantom{\rule{0ex}{0ex}}$

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