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

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

Steps

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

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

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

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

Find f such that f′(x)=4x2+5x−5 and f(0)=3f(x)=

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