Q 31 - Calculus

Question

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

$f^{\text{'}}\left(x\right)=8e^{4x}+1,f\left(0\right)=3$

Step-by-Step Solution

Answer

The anti derivative is $f\left(x\right)=2e^{4x}+x+1$

Step 1:Given informatiom

Given expression is $f\text{'}\left(x\right)=8e^{4x}+1,f\left(0\right)=3$

Step 2:Simplification

Given $f\text{'}\left(x\right)=8e^{4x}+1$

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

here $e^{4x}$ is derivative of $\frac{e^{4x}}{4}$

Therefore for $8e^{4x}$ antiderivative is $2e^{4x}$

And 1 is derivative of x

Hence

width="127" height="23" style="max-width: none; vertical-align: -5px;" $f\left(x\right)=2e^{4x}+x+c$

Now,

$$\begin{gathered} f\left(0\right)=3 \\ \Rightarrow 3=2e^0+0+c \\ \Rightarrow 3=2+c \\ \Rightarrow c=1 \end{gathered}$$

therefore,

$f\left(x\right)=2e^{4x}+x+1$