If \(f ' '(x) = 2\), \(f ' (1) = 0\) and \(f(0) = -

Explanation

\(f ' ' (x) = \frac{\mathrm d ^{2} y}{\mathrm d ^{2} x} = 2\)

\(f ' (x) = \int 2 \mathrm {d} x\)

= \(2x + c\)

When x = 1, f'(x) = 0.

\(2(1) + c = 0 \implies c = -2\)

\(\therefore f ' (x) = 2x - 2\)

\(f(x) = \int (2x - 2) \mathrm {d} x\)

= \(x^{2} - 2x + c\)

When x = 0, f(x) = -8

\(0^{2} - 2(0) + c = -8\)

\(c = -8\)

\(\therefore f(x) = x^{2} - 2x - 8\)