Q.Find the intervals of monotonicities and hence find the local extremum for the following functions:

Step 1 (i). $f(x)=2x^3+3x^2-12x$.

$f'(x)=6x^2+6x-12=6(x+2)(x-1)$. Critical numbers $x=-2,1$.

For $x<-2$: both factors negative, product positive $\Rightarrow f'>0$ (increasing).

For $-2<x<1$: $(x+2)>0,(x-1)<0\Rightarrow f'<0$ (decreasing).

For $x>1$: both positive $\Rightarrow f'>0$ (increasing).

$f(-2)=-16+12+24=20$: local maximum. $f(1)=2+3-12=-7$: local minimum.

Step 2 (ii). $f(x)=\dfrac{x}{x-5}$.

This is negative for every $x\ne5$ (domain excludes $x=5$), so $f$ is strictly decreasing on $(-\infty,5)$ and on $(5,\infty)$. Since $f'$ is never zero, there are no local extrema.

Step 3 (iii). $f(x)=\dfrac{e^x}{1-e^x}$.

Always positive (domain excludes $x=0$ where $1-e^x=0$), so $f$ is strictly increasing on $(-\infty,0)$ and on $(0,\infty)$. No local extrema.

Step 4 (iv). $f(x)=\dfrac{x^3}{3}-\log x$, $x>0$.

$f'(x)=x^2-\dfrac1x=\dfrac{x^3-1}{x}$. Critical number: $x^3=1\Rightarrow x=1$ (only positive real root).

For $0<x<1$: $x^3-1<0$, $x>0\Rightarrow f'<0$ (decreasing). For $x>1$: $f'>0$ (increasing).

Local minimum at $x=1$: $f(1)=\tfrac13-0=\tfrac13$.

Step 5 (v). $f(x)=\sin x\cos x+5=\tfrac12\sin2x+5$, $x\in(0,2\pi)$.

$f'(x)=\cos2x$. Zero when $2x=\tfrac{\pi}2,\tfrac{3\pi}2,\tfrac{5\pi}2,\tfrac{7\pi}2\Rightarrow x=\tfrac{\pi}4,\tfrac{3\pi}4,\tfrac{5\pi}4,\tfrac{7\pi}4$. …

• Treating $x=5$ or $x=0$ (points outside the domain) as genuine critical numbers requiring classification …