If f'(x) = x^2 + 5 and f(0) = - 1, then f(x) = - Vidyadip

MCQ

If $f'(x) = {x^2} + 5$ and $f(0) = - 1$, then $f(x) = $

A
${x^3} + 5x - 1$
B
${x^3} + 5x + 1$
✓
$\frac{1}{3}{x^3} + 5x - 1$
D
$\frac{1}{3}{x^3} + 5x + 1$

✓

Answer

Correct option: C.

$\frac{1}{3}{x^3} + 5x - 1$

c
(c) Given that $f'(x) = {x^2} + 5$ and $f(0) = - 1$
==> $f(x) = \frac{{{x^3}}}{3} + 5x + c$. If $x = 0,$ then $f(0) = c$

==> $c = - 1$.
Hence $f(x) = \frac{{{x^3}}}{3} + 5x - 1.$

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