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

(a) ${x^2} + 2x - 5$

(b) $10 - 6x - 2{x^2}$

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

(d) $6 - 9x - {x^2}$

(e) ${\left( {x + 1} \right)^3}{\left( {x - 3} \right)^3}$

(a) We have, $f(x) = {x^2} + 2x - 5$ …(i)
$f(x)$ being a polynomial, is continuous and derivable on s$R$.

Differentiating (i), w.r.t. $x$, we get, $f'(x) = 2x + 2$
For function to be increasing, $f'(x) > 0$
$\Rightarrow 2x + 2 > 0 \Rightarrow x > - 1$

For function to be decreasing, $f'(x) < 0$
$\Rightarrow 2x + 2 < 0 \Rightarrow x < - 1$
Therefore, $f(x)$ is strictly increasing for $x > - 1.$

$f(x)$ is strictly decreasing for $x < - 1.$

(b) We have, $f(x) = 10 - 6x - 2{x^2}$ …(i)

$f(x)$ being a polynomial, is continuous and derivable on $R$

Differentiating (i) w.r.t. $x$, we get
$f(x) = 0 - 6 - 2(2x) = - 6 - 4x$

For function to be increasing,
$f(x) > 0 \Rightarrow - 6 - 4x > 0 \Rightarrow - 4x > 6$
$\Rightarrow x < - \cfrac{3}{2} \Rightarrow x \in \left( { - \infty , - \cfrac{3}{2}} \right)$

Therefore, $f(x)$ is strictly increasing if $x < - \cfrac{3}{2}.$

For function to be decreasing,
$f(x) < 0 \Rightarrow - 6 - 4x < 0 \Rightarrow - 4x < 6$
$\Rightarrow x > - \cfrac{3}{2} \Rightarrow x \in \left( { - \cfrac{3}{2},\infty } \right)$

$\Rightarrow f(x)$ is strictly decreasing if $x > - \cfrac{3}{2}.$

(c) We have, $f(x) = - 2{x^3} - 9{x^2} - 12x + 1$ …(i)
$f\left( x \right)$ being a polynomial, is continuous and derivable on $R$

Differentiating (i) w.r.t. $x$, we get
$f(x) = - 2(3{x^2}) - 9(2x\rangle - 12$
$= - 6{x^2} - 18x - 12 = - 6({x^2} + 3x + 2) = - 6(x + 1)(x + 2)$

For critical points, putting $f'(x) = 0$
$\Rightarrow - 6\left( {x + 1} \right)\left( {x + 2} \right) = 0 \Rightarrow x = - 1,x = - 2$

The points $x = - 1, - 2$ divide the real line into following disjoint intervals $( - \infty ,\; - 2),( - 2,\; - 1)$ and $( - 1,\;\infty )$
Interval

Sign of
$f\prime \left( x \right) = - 6\left( {x + 1} \right)\left( {x + 2} \right)$

Nature of function
$\left( { - \infty , - 2} \right)$
$\left( { - ve} \right)\left( { - ve} \right)\left( { - ve} \right) = - ve$

$f$ is strictly decreasing
$\left( { - 2, - 1} \right)$
$\left( { - ve} \right)\left( { - ve} \right)\left( { + ve} \right) = + ve$

$f$ is strictly increasing
$\left( { - 1,\infty } \right)$
$\left( { - ve} \right)\left( { + ve} \right)\left( { + ve} \right) = - ve$

$f$ is strictly decreasing

Hence, $f$ is strictly increasing on $\left( { - 2, - 1} \right)$ and strictly decreasing in the interval $\left( { - \infty , - 2} \right) \cup \left( { - 1,\infty } \right)$

(d) We have, $f\left( x \right) = 6 - 9x - {x^2}$ …(i)
$f\left( x \right)$ being a polynomial, is continuous and derivable on $R$

Differentiating (i) w.r.t. $x$, we get
$f\prime \left( x \right) = - 9 - 2x$
For function to be increasing,
$f\prime \left( x \right) > 0 \Rightarrow - 9 - 2x > 0 \Rightarrow - 2x > 9$
$\Rightarrow$ $x < - \cfrac{9}{2} \Rightarrow x \in \left( { - \infty , - \cfrac{9}{2}} \right)$
Therefore, $f$ is strictly decreasing on $\left( { - \cfrac{9}{2},\infty } \right)$.

(e) We have, $f\left( x \right) = {\left( {x + 1} \right)^3}{\left( {x - 3} \right)^3}$
$f\left( x \right)$ being a polynomial, is continuous and derivable on $R$

Differentiating (i) w.r.t. $x$, we get
$f\prime \left( x \right) = {\left( {x + 1} \right)^3} \cdot 3{\left( {x - 3} \right)^2} \cdot 1 + {\left( {x - 3} \right)^3} \cdot 3{\left( {x + 1} \right)^2} \cdot 1$

$= {\left( {x - 3} \right)^2}{\left( {x + 1} \right)^2}\left\{ {3\left( {x + 1} \right) + 3\left( {x - 3} \right)} \right\}$

${\left( {x - 3} \right)^2}{\left( {x + 1} \right)^2}\left\{ {6x - 6} \right\} = 6{\left( {x - 3} \right)^2}{\left( {x + 1} \right)^2}\left( {x - 1} \right)$

For critical points, put $f\prime \left( x \right) = 0$
$6{\left( {x - 3} \right)^2}{\left( {x + 1} \right)^2}\left( {x - 1} \right) = 0 \Rightarrow x = 3,x = - 1,x = 1$

The above points divide the real line into following disjoint intervals
$\left( { - \infty , - 1} \right),\left( { - 1,1} \right),\left( {1,3} \right)\,\,{\rm{and}}\,\left( {3,\infty } \right)$
Interval
Sign of
$f\prime \left( x \right) = 6{\left( {x - 3} \right)^2}{\left( {x + 1} \right)^2}\left( {x - 1} \right)$

Nature of function

$f$
$\left( { - \infty , - 1} \right)$
$\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right)\left( { - ve} \right) = - ve$
$f$ is strictly decreasing

$\left( { - 1,1} \right)$
$\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right)\left( { - ve} \right) = - ve$

$f$ is strictly decreasing
$\left( {1,3} \right)$
$\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right) = + ve$

$f$ is strictly increasing
$\left( {3,\infty } \right)$
$\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right) = + ve$

$f$ is strictly increasing

Thus, $f$ is strictly increasing on $\left( {1,3} \right) \cup \left( {3,\infty } \right)$ and strictly decreasing on $\left( { - \infty , - 1} \right) \cup \left( { - 1,1} \right)$