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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