If f'(x) = 8x^3 - 2x, f'(2) = 8, find f'(x). - Vidyadip

Question

If $f'(x) = 8x^3 - 2x, f'(2) = 8$, find $f'(x)$.

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Answer

We have,
$\text{f}'\text{(x)}=8\text{x}^3-2\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\int\text{f}'\text{(x)dx}=\int(8\text{x}^3-2\text{x})\text{dx}$
$\Rightarrow\text{f}'\text{(x)}=\int(8\text{x}^3-2\text{x})\text{dx}$
$=\int8\text{x}^3\text{dx}-\int2\text{ x dx}$
$=\frac{8\text{x}^4}{4}-\frac{2\text{x}^2}{2}+\text{C}$
$=2\text{x}^4-\text{x}^2+\text{C}$
$\Rightarrow\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2+\text{C}\ \dots(1)$
Since, $\text{f}'(2)=8$
$\therefore\ \text{f}'(2)=2(2)^4-(2)^2+\text{C}=8$
$\Rightarrow32-4+\text{C}=8$
$\Rightarrow28+\text{C}=8$
$\Rightarrow\text{C}=-20$
Putting C = -20 in eq. (1), we get
$\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2-20$
Hence, $\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2-20$

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