Question

Let $f\left(x\right)=e^x-e^{-x},g\left(x\right)=\ln (x^2-5x+6),$ then which of the following(s) is(are) correct about $h\left(x\right)=f\left(g\right(x\left)\right)$

• A. increasing in $(2,\infty )$
• B. increasing in $(3,\infty )$
• C. decreasing in $(-\infty ,2)$
• D. decreasing in $(-\infty ,\frac{5}{2})$

Answer 1

Answer: (B), (C)

decreasing in $(-\infty ,2)$

Given : $f\left(x\right)=e^x-e^{-x},g\left(x\right)=\ln (x^2-5x+6)$
we can see $f\left(x\right)$ is defined for all $x\in R$ and $g\left(x\right)$ is defined when $x^2-5x+6>0\Rightarrow x\in (-\infty ,2)\cup (3,\infty )\cdots \left(1\right)$
Now, $h\left(x\right)=f\left(g\right(x\left)\right)=x^2-5x+6-\frac{1}{x^2-5x+6}$
$(\because e^{\ln y}=y,y>0)$
$\Rightarrow h^{\prime }\left(x\right)=2x-5+\frac{2x-5}{(x^2-5x+6{)}^2}$
$\Rightarrow h^{\prime }\left(x\right)=(2x-5)(1+\frac{1}{(x^2-5x+6{)}^2})$
$\because (1+\frac{1}{(x^2-5x+6{)}^2})>0$ for all $x$ in its domain.
For increasing, $h^{\prime }\left(x\right)\ge 0:$
$\Rightarrow x\ge \frac{5}{2}$
from $\left(i\right),x\in (3,\infty )$
For decreasing, $h^{\prime }\left(x\right)\le 0:$
$\Rightarrow x\le \frac{5}{2}$
from $\left(i\right),x\in (-\infty ,2)$