Q. 3 - Calculus

Question

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

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

Step-by-Step Solution

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 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

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

Step 3. Finding the value f(c)

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