Question

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

$f\text{'}\left(x\right)=\frac{1}{1+4x},f\left(0\right)=1$

Short answer

The antiderivative is$f\left(x\right)=\log (1+4x)+1$

Step-by-step solution

Step 1:Given information

The given expression is$f\text{'}\left(x\right)=\frac{1}{1+4x},f\left(0\right)=1$

Step 2:Simplificaion

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

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

here $\frac{1}{x}$is the derivative of $\log x$where$\log \left(x\right)=1+4x$

Therefore $\frac{1}{1+4x}$is the derivative of $\log \left(1+4x\right)$

hence,

$f\left(x\right)=\log \left(1+4x\right)$$+c$

now,

$$\begin{gathered} f\left(0\right)=1 \\ \Rightarrow 1=\log \left(1+0\right)+c \\ \Rightarrow c=1 \end{gathered}$$

Therefore antiderivative is$f\left(x\right)=\log (1+4x)+1$