find f such that f(x)=5x² + 5x - 4 and f(0)=2. f(x)=

Question

Accepted Answer

# Explanation:
## Step1: Integrate $f^{\prime}(x)$
We know that if $f^{\prime}(x)=5x^{2}+5x - 4$, then by the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$):
$$
\begin{align*}
f(x)&=\int(5x^{2}+5x - 4)dx\
&=5\int x^{2}dx+5\int xdx-4\int dx\
&=5\times\frac{x^{3}}{3}+5\times\frac{x^{2}}{2}-4x + C\
&=\frac{5}{3}x^{3}+\frac{5}{2}x^{2}-4x + C
\end{align*}
$$
## Step2: Use the initial condition $f(0) = 2$
Substitute $x = 0$ into $f(x)=\frac{5}{3}x^{3}+\frac{5}{2}x^{2}-4x + C$.
When $x = 0$, $f(0)=\frac{5}{3}(0)^{3}+\frac{5}{2}(0)^{2}-4(0)+C$.
Since $f(0) = 2$, we have $C = 2$.

# Answer:
$f(x)=\frac{5}{3}x^{3}+\frac{5}{2}x^{2}-4x + 2$

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QUESTION IMAGE

Question

Explanation:

Step1: Integrate \(f^{\prime}(x)\)

Step2: Use the initial condition \(f(0) = 2\)

Answer: