find f.\n\n$f(x)=-2 + 24x - 12x^{2}$, $f(0)=6$, $f(0)=14$\n\n$f(x)=$

Explanation:

Step1: Integrate (f''(x)) to find (f'(x))

Integrate term - by - term: [ \begin{align*} f'(x)&=\int(-2 + 24x-12x^{2})dx\ &=\int(-2)dx+\int24xdx-\int12x^{2}dx\ &=-2x + 24\times\frac{x^{2}}{2}-12\times\frac{x^{3}}{3}+C\ &=-2x + 12x^{2}-4x^{3}+C \end{align*} ] Use the initial condition (f'(0) = 14). Substitute (x = 0) into (f'(x)): (f'(0)=-2\times0+12\times0^{2}-4\times0^{3}+C), so (C = 14). Then (f'(x)=-4x^{3}+12x^{2}-2x + 14)

Step2: Integrate (f'(x)) to find (f(x))

Integrate term - by - term: [ \begin{align*} f(x)&=\int(-4x^{3}+12x^{2}-2x + 14)dx\ &=\int(-4x^{3})dx+\int12x^{2}dx-\int2xdx+\int14dx\ &=-4\times\frac{x^{4}}{4}+12\times\frac{x^{3}}{3}-2\times\frac{x^{2}}{2}+14x+D\ &=-x^{4}+4x^{3}-x^{2}+14x+D \end{align*} ] Use the initial condition (f(0)=6). Substitute (x = 0) into (f(x)): (f(0)=-0^{4}+4\times0^{3}-0^{2}+14\times0+D), so (D = 6)

Answer:

(f(x)=-x^{4}+4x^{3}-x^{2}+14x + 6)