Q. 76 - Calculus

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   

$f\text{'}\left(x\right)=x(4-2x) ; f\left(0\right)=0$

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 2: Find the antiderivative

We get,

$f\text{'}\left(x\right)=4x-2x^2$

We know that differentiating a power function decreases the power by one we can start with the function $f\left(x\right)=x^2-x^3$

The derivative can be given as

$f\text{'}\left(x\right)=2x-3x^2$

Which is nearly equal now we just adjust the coefficients

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

We are also given 

$$\begin{gathered} f\left(0\right)=0 \\ Substituting we get \\ c=0 \end{gathered}$$

Hence the antiderivative can be given as 

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

Step 3: Conclusion

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