Question
Find intervals in which the function given by $f(x) = \frac{3}{{10}}{x^4} - \frac{4}{5}{x^3} - 3{x^2} + \frac{{36}}{5}x + 11$ is increasing or decreasing.
✓
Answer
$f(x) = \frac{3}{{10}}{x^4} - \frac{4}{5}{x^3} - 3{x^2} + \frac{{36}}{5}x + 11$
$f'(x) = \frac{3}{{10}}.4{x^3} - \frac{4}{5}3{x^2} - 6x + \frac{{36}}{5}$
$ = \frac{6}{5}{x^3} - \frac{{12}}{5}{x^2} - \frac{{6x}}{1} + \frac{{36}}{5}$
$f'(x) = \frac{{6{x^3} - 12{x^2} - 30x + 36}}{5}$
$ = \frac{6}{5}\left[ {{x^3} - 2{x^2} - 5x + 6} \right]$
let f'(x) = 0
$\frac{6}{5}({x^3} - 2{x^2} - 5x + 6) = 0$
${x^3} - 2{x^2} - 5x + 6 = 0$
Put x = 1
f'(1) = 1 - 2 - 5 + 6 = 0
x-1 is the factor of f'(x)
$f'(x) = \frac{6}{5}(x - 1)({x^2} - x - 6)$
$ = \frac{6}{5}(x - 1)\left[ {{x^2} - 3x + 2x - 6} \right]$
$ = \frac{6}{5}(x - 1)\left[ {x(x - 3) + 2(x - 2)} \right]$
$ = \frac{6}{5}(x - 1)(x - 3)(x + 2)$
put f'(x) = 0
x = 1, x = 3, x = -2
$f'(x) = \frac{3}{{10}}.4{x^3} - \frac{4}{5}3{x^2} - 6x + \frac{{36}}{5}$
$ = \frac{6}{5}{x^3} - \frac{{12}}{5}{x^2} - \frac{{6x}}{1} + \frac{{36}}{5}$
$f'(x) = \frac{{6{x^3} - 12{x^2} - 30x + 36}}{5}$
$ = \frac{6}{5}\left[ {{x^3} - 2{x^2} - 5x + 6} \right]$
let f'(x) = 0
$\frac{6}{5}({x^3} - 2{x^2} - 5x + 6) = 0$
${x^3} - 2{x^2} - 5x + 6 = 0$
Put x = 1
f'(1) = 1 - 2 - 5 + 6 = 0
x-1 is the factor of f'(x)
$f'(x) = \frac{6}{5}(x - 1)({x^2} - x - 6)$
$ = \frac{6}{5}(x - 1)\left[ {{x^2} - 3x + 2x - 6} \right]$
$ = \frac{6}{5}(x - 1)\left[ {x(x - 3) + 2(x - 2)} \right]$
$ = \frac{6}{5}(x - 1)(x - 3)(x + 2)$
put f'(x) = 0
x = 1, x = 3, x = -2
| int | Sign of f’(x) | Result |
|---|---|---|
| $( - \infty , - 2)$ | -ve | Decrease |
| (-2, 1) | +ve | Increase |
| (1,3) | -ve | Decrease |
| $(3,\infty )$ | +tve | increase |
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