find f.

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

$f(x)=$

Question

find f.

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

$f(x)=$

Accepted Answer

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

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QUESTION IMAGE

Question

Explanation:

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

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

Answer: