Question
Find the interval of the function that is strictly increasing or decreasing: $(x + 1)^3 (x - 3)^{3 }$
✓
Answer
$f(x) = {(x + 1)^3}{(x - 3)^3}$
$f\ '(x) = {(x + 1)^3}.3{(x - 3)^2} + {(x - 3)^3}.3{(x + 1)^2}$
$ = 3{(x + 1)^2}{(x - 3)^2}[x + 1 + x - 3]$
$ = 3{(x + 1)^2}{(x - 3)^2}[2x - 2]$
$ = 6{(x + 1)^2}{(x - 3)^2}(x - 1)$
Put $f\ '(x) = 0$
$x = -1, 3, 1$
$f\ '(x) = {(x + 1)^3}.3{(x - 3)^2} + {(x - 3)^3}.3{(x + 1)^2}$
$ = 3{(x + 1)^2}{(x - 3)^2}[x + 1 + x - 3]$
$ = 3{(x + 1)^2}{(x - 3)^2}[2x - 2]$
$ = 6{(x + 1)^2}{(x - 3)^2}(x - 1)$
Put $f\ '(x) = 0$
$x = -1, 3, 1$
| int | Sign of $f\ ’(x)$ | Result |
| $( - \infty , - 1)$ | $-ve$ | Decrease |
| $(-1, 1)$ | $-ve$ | Decrease |
| $(1, 3)$ | $+ve$ | Increase |
| $(3,\infty )$ | $+ve$ | Increase |
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