Question

Find a function f that has the given derivative f' and value f(c), if possible.

$f\text{'}\left(x\right)=3e^{2x}+1,f\left(0\right)=2$

Short answer

$\mathrm{Hence}\mathrm{the}\mathrm{function}\mathrm{is}3{\mathrm{e}}^{2\mathrm{x}}\times 2+\mathrm{x}-4$

The value of f(-4) =$3\times e^{-8}-8$

Step-by-step solution

Step 1. Given 

The given function is$f\text{'}\left(x\right)=3e^{2x}+1,f\left(0\right)=2$

Step 2. Finding the function using integration 

$$\begin{gathered} \mathrm{Integrating}\mathrm{function}\mathrm{with}\mathrm{respect}\mathrm{to}x \\ \int f\text{'}\left(x\right)=\int 3e^{2x}+1,f\left(0\right)=2 \\ =\int 3e^{2x}+1dx \\ =3{\mathrm{e}}^{2\mathrm{x}}\times 2+\mathrm{x}+\mathrm{c} \\ \mathrm{where}\mathrm{c}\mathrm{is}\mathrm{constant} \\ \mathrm{when}\mathrm{f}\left(0\right)=2 \\ \mathrm{f}\left(0\right)=3{\mathrm{e}}^{2*0}*2+0+\mathrm{c} \\ 2=6+\mathrm{c} \\ \mathrm{c}=-4 \\ \mathrm{Hence}\mathrm{the}\mathrm{function}\mathrm{is}3{\mathrm{e}}^{2\mathrm{x}}\times 2+\mathrm{x}-4 \end{gathered}$$

Step 3. Finding the value f(c)

$$\begin{gathered} f\left(x\right)=3{\mathrm{e}}^{2\mathrm{x}}\times 2+\mathrm{x}-4 \\ f(-4)=3\times e^{2*-4}+(-4)-4 \\ =3\times e^{-8}-8 \end{gathered}$$