Question

Find

a function $f$that has the given derivative $f$$\text{'}$ and value f(c).

$f\text{'}\left(x\right)=x\left(3x+1\right),f\left(2\right)=4$

Short answer

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

Step-by-step solution

Step 1:Given information

The given expression is$f\text{'}\left(x\right)=x\left(3x+1\right)$$,f\left(2\right)=4$

Step 2:Simplification

given

$$\begin{gathered} f\text{'}\left(x\right)=x\left(3x+1\right) \\ =3x^2+x \end{gathered}$$

We will use an intelligent guess-and-check method to find f

Here $3x^2$is derivative of $x^3$and x is derivative of $\frac{x^2}{2}$

hence,

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

Now,$f\left(2\right)=4$

$\Rightarrow 4={\left(2\right)}^3+\frac{2^2}{2}+c$

$$\begin{gathered} \Rightarrow 4=8+2+c \\ \Rightarrow c=-6 \end{gathered}$$

Therefore

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