Find the intervals in which the following functions are strictly increasing or decreasing:

Solution:

(a) We have,

$f(x)=x^{2}+2 x-5$

$\therefore f^{\prime}(x)=2 x+2$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1$

Point $x=-1$ divides the real line into two disjoint intervals i.e., $(-\infty,-1)$ and $(-1, \infty)$.

In interval $(-\infty,-1), f^{\prime}(x)=2 x+2<0$

$\therefore f$ is strictly decreasing in interval $(-\infty,-1)$.

Thus, $f$ is strictly decreasing for $x<-1$.

In interval $(-1, \infty), f^{\prime}(x)=2 x+2>0$

$\therefore f$ is strictly increasing in interval $(-1, \infty)$.

Thus, $f$ is strictly increasing for $x>-1$.

(b) We have,

$f(x)=10-6 x-2 x^{2}$

$\therefore f^{\prime}(x)=-6-4 x$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-\frac{3}{2}$

The point $x=-\frac{3}{2}$ divides the real line into two disjoint intervals i.e., $\left(-\infty,-\frac{3}{2}\right)$ and $\left(-\frac{3}{2}, \infty\right)$.

In interval $\left(-\infty,-\frac{3}{2}\right)$ i.e., when $x<-\frac{3}{2}, f^{\prime}(x)=-6-4 x>0$.

$\therefore f$ is strictly increasing for $x<-\frac{3}{2}$.

In interval $\left(-\frac{3}{2}, \infty\right)$ i.e., when $x>-\frac{3}{2}, f^{\prime}(x)=-6-4 x<0$.

$\therefore f$ is strictly decreasing for $x>-\frac{3}{2}$.

(c) We have,

$f(x)=-2 x^{3}-9 x^{2}-12 x+1$

$\therefore f^{\prime}(x)=-6 x^{2}-18 x-12=-6\left(x^{2}+3 x+2\right)=-6(x+1)(x+2)$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1$ and $x=-2$

Points $x=-1$ and $x=-2$ divide the real line into three disjoint intervals i.e., $(-\infty,-2),(-2,-1)$, and $(-1, \infty)$.

In intervals $(-\infty,-2)$ and $(-1, \infty)$ i.e., when $x<-2$ and $x>-1$,

$f^{\prime}(x)=-6(x+1)(x+2)<0$

$\therefore f$ is strictly decreasing for $x<-2$ and $x>-1$.

Now, in interval $(-2,-1)$ i.e., when $-20$.

$\therefore f$ is strictly increasing for $-2

(d) We have,

$f(x)=6-9 x-x^{2}$

$\therefore f^{\prime}(x)=-9-2 x$

Now, $f^{\prime}$

$(x)=0$ gives $x=-\frac{9}{2}$

The point $x=-\frac{9}{2}$ divides the real line into two disjoint intervals i.e., $\left(-\infty,-\frac{9}{2}\right)$ and $\left(-\frac{9}{2}, \infty\right)$.

In interval $\left(-\infty,-\frac{9}{2}\right)$ i.e., for $x<-\frac{9}{2}, f^{\prime}(x)=-9-2 x>0$.

$\therefore f$ is strictly increasing for $x<-\frac{9}{2}$.

In interval $\left(-\frac{9}{2}, \infty\right)$ i.e., for $x>-\frac{9}{2}, f^{\prime}(x)=-9-2 x<0$.

$\therefore f$ is strictly decreasing for $x>-\frac{9}{2}$.

(e) We have,

$f(x)=(x+1)^{3}(x-3)^{3}$

$\begin{aligned} f^{\prime}(x) &=3(x+1)^{2}(x-3)^{3}+3(x-3)^{2}(x+1)^{3} \\ &=3(x+1)^{2}(x-3)^{2}[x-3+x+1] \\ &=3(x+1)^{2}(x-3)^{2}(2 x-2) \\ &=6(x+1)^{2}(x-3)^{2}(x-1) \end{aligned}$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1,3,1$

The points $x=-1, x=1$, and $x=3$ divide the real line into four disjoint intervals i.e., $(-\infty,-1),(-1,1),(1,3)$, and $(3, \infty)$.

In intervals $(-\infty,-1)$ and $(-1,1), f^{\prime}(x)=6(x+1)^{2}(x-3)^{2}(x-1)<0$.

$\therefore f$ is strictly decreasing in intervals $(-\infty,-1)$ and $(-1,1)$.

In intervals $(1,3)$ and $(3, \infty), f^{\prime}(x)=6(x+1)^{2}(x-3)^{2}(x-1)>0$

$\therefore f$ is strictly increasing in intervals $(1,3)$ and $(3, \infty)$.