Question

Find the interval of increase and decrease of the following functions.
$f\left(x\right)=x+\ln (1-4x)$

Answer 1

$f\left(x\right)=x+\ln (1-4x)$

First, $f^{\prime }\left(x\right)=1+\frac{1}{1-4x}(-4)$

$\Rightarrow 1-\frac{4}{1-4x}=0$

$\Rightarrow 1-4x-4=0$

$\Rightarrow x=\frac{-3}{4}$

Critical point $=\frac{-3}{4},\frac{1}{4}$

Also, $f\left(x\right)$ is undefined at $x=\frac{1}{4}$ since

$\ln 0$ would then be undefined.

domain of $f\left(x\right)$ is $x<\frac{1}{4}$, because for $x>\frac{1}{4},(1-4x)$ will be negative and then $ln(1-4x)$ will not be defined.

The function montone intervals are

$-\infty <x<\frac{-3}{4},\frac{-3}{4}<x<\frac{1}{4}$

Evaluate the derivative at a point on the interval $-\infty <x<\frac{-3}{4}$

$\Rightarrow f^{\prime }\left(x\right)=1-\frac{4}{1-4x}>0$

$\therefore f\left(x\right)$ is increasing on $-\infty <x<\frac{-3}{4}$

and similarly, since $f^{\prime }\left(x\right)<0$ on $\frac{-3}{4}<x<\frac{1}{4}$

$\therefore f\left(x\right)$ is decreasing on $\frac{-3}{4}<x<\frac{1}{4}$.