find f such that $f(x)=7x^2 + 5x - 7$ and $f(0)=9$.

$f(x)=\square$

Question

find f such that $f(x)=7x^2 + 5x - 7$ and $f(0)=9$.

$f(x)=\square$

Accepted Answer

# Explanation:
## Step1: Integrate $ f'(x) $
To find $ f(x) $, we integrate $ f'(x) = 7x^2 + 5x - 7 $. Using the power rule for integration $ \int x^n dx=\frac{x^{n + 1}}{n+1}+C $ (for $ n
eq - 1$):
$$
\begin{align*}
f(x)&=\int(7x^{2}+5x - 7)dx\
&=7\int x^{2}dx+5\int xdx-7\int dx\
&=7\times\frac{x^{3}}{3}+5\times\frac{x^{2}}{2}-7x + C\
&=\frac{7}{3}x^{3}+\frac{5}{2}x^{2}-7x + C
\end{align*}
$$
## Step2: Find the constant $ C $
We know that $ f(0)=9 $. Substitute $ x = 0 $ into $ f(x) $:
$$
\begin{align*}
f(0)&=\frac{7}{3}(0)^{3}+\frac{5}{2}(0)^{2}-7(0)+C\
9&=0 + 0-0 + C
\end{align*}
$$
So $ C = 9 $.
## Step3: Write the final function $ f(x) $
Substitute $ C = 9 $ back into the expression for $ f(x) $:
$$
f(x)=\frac{7}{3}x^{3}+\frac{5}{2}x^{2}-7x + 9
$$

# Answer:
$ \frac{7}{3}x^{3}+\frac{5}{2}x^{2}-7x + 9 $

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QUESTION IMAGE

Question

Explanation:

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

Step2: Find the constant \( C \)

Step3: Write the final function \( f(x) \)

Answer: