Answer
$f(x) = \frac{2}{5}x^5 + \frac{5}{2} x^2 + x -3.9$
Work Step by Step
$f'(x) = 2x^4 + 5x + C$
$f'(1) = 2(1)^4 + 5(1) + C$
Given $f'(1) = 8$
$8 = 2 + 5 + C$
$C = 1$
$f'(x) = 2x^4 + 5x + 1$
Find $f$:
$f(x) = 2 \times \frac{x^{4+1}}{4+1} + 5 \times \frac{x^{1+1}}{1+1} + 1(x) $
$f(x) = 2 \times \frac{x^5}{5} + 5 \times \frac{x^2}{2} + x $
$f(x) = \frac{2x^5}{5} + \frac{5x^2}{2} + x + C$
$f(1) = \frac{2(1)^5}{5} + \frac{5(1)^2}{2} + 1 + C$
$f(1) = \frac{2}{5} + \frac{5}{2} + 1 + C$
Given $f(1) = 0$:
$0 = \frac{39}{10} + C$
$C = -3.9$
$f(x) = \frac{2}{5}x^5 + \frac{5}{2} x^2 + x -3.9$
