Problem

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

Solution

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

From Solvely APP
Source: https://solvelyapp.com/problems/7934/

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