Find $f$ such that $f^{\prime}(x)=4 x^{2}+5 x-5$ and $f(0)=3$
\[
f(x)=
\]

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Answer

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

Steps

Step 1 ：Given the derivative function $f'(x) = 4x^2 + 5x - 5$ and the initial condition $f(0) = 3$.

Step 2 ：Integrate the derivative function to find $f(x)$: $f(x) = \int f'(x) dx = \int (4x^2 + 5x - 5) dx = \frac{4}{3}x^3 + \frac{5}{2}x^2 - 5x + C$.

Step 3 ：Substitute the initial condition $f(0) = 3$ into the function to find the constant $C$: $3 = \frac{4}{3}(0)^3 + \frac{5}{2}(0)^2 - 5(0) + C$, so $C = 3$.

Step 4 ：Substitute $C = 3$ back into the function to get the final answer: $f(x) = \frac{4}{3}x^3 + \frac{5}{2}x^2 - 5x + 3$.

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

Find $f$ such that $f^{\prime}(x)=4 x^{2}+5 x-5$ and $f(0)=3$\[f(x)=\]

Answer

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Find $f$ such that $f^{\prime}(x)=4 x^{2}+5 x-5$ and $f(0)=3$
\[
f(x)=
\]