Question
If f'(x) = x + b, f'(1) = 5, f'(2) = 13, find f'(x).

Answer

$\text{f}'\text{(x)}=\text{x + b},\text{f}'(1)=5,\text{f}'(2)=13$
$\text{f}'\text{(x)}=\text{x + b}$
$\int\text{f}'\text{(x) dx}=\int(\text{x + b})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\text{bx}+\text{C}\ \dots(1)$
$\text{f}'(1)=5,\text{f}'(2)=13$ (Given)
putting x = 1 in (1)
$\text{f}'(1)=\frac{1^2}{2}+\text{b}_1+\text{C}$
$5=\frac{1}{2}+\text{b + C}\ \dots(2)$
Putting x = 2 in (1)
$\text{f}'(2)=\frac{2^2}{2}+\text{b}_2+\text{C}$
$13=\frac{4}{2}+2\text{b + C}$
$13=2+2\text{b + C}\ \dots(3)$
Solving (2) and (3) we get,
$\text{b}=\frac{13}{2}\text{ and }\text{C}=-2$
Thus, $\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{13}{2}\text{x}-2$

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