Problem

Find $f$ such that $f^{\prime}(x)=x^{2}+2$ and $f(0)=8$ \[ f(x)= \]

Solution

Step 1 :The problem is asking for a function \(f(x)\) such that its derivative is \(x^{2}+2\) and the function evaluated at 0 is 8. To find such a function, we need to integrate the derivative function and then adjust the constant of integration so that \(f(0)=8\).

Step 2 :Integrating the derivative function \(x^{2}+2\) gives us \(\frac{x^{3}}{3}+2x+C\), where \(C\) is the constant of integration.

Step 3 :We then set \(f(0)=8\) to find the value of \(C\). Substituting \(0\) into the function gives us \(C=8\).

Step 4 :Substituting \(C=8\) back into the function gives us the final function \(f(x)=\frac{x^{3}}{3}+2x+8\).

Step 5 :\(\boxed{f(x)=\frac{x^{3}}{3}+2x+8}\) is the function that satisfies the given conditions.

From Solvely APP
Source: https://solvelyapp.com/problems/8417/

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