Question

Find a function that has the given derivative and value. In each case you can find the answer with an educated guess and check process it may be helpful to do some preliminary algebra

$f\text{'}\left(x\right)=3x^5-2x^2+4;f\left(0\right)=1$

Short answer

The antiderivative can be given as$f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+1$

Step-by-step solution

Given information

We are given the derivative as$f\text{'}\left(x\right)=3x^5-2x^2+4;f\left(0\right)=1$

Find the antiderivative

We know that differentiating a power function decreases the power by one we can start with the function $f\left(x\right)=x^6-x^3+4x+c$

On differentiating the above function we get,

$f\text{'}\left(x\right)=6x^5-3x^2+4$which is nearly equal

Now we only adjust the coefficient

We get, $f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+c$

We are also given that $$\begin{gathered} f\left(0\right)=1 \\ Substittuingthisinthefunctionweget, \\ f\left(0\right)=c \\ c=1 \end{gathered}$$

Hence the antiderivative becomes

$f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+1$

Conclusion

The antiderivative can be given as$f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+1$