Question
Find the intervals in which the function f given by $f(x) = 2x^3 – 3x^2 – 36x + 7$ is increasing.
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Answer
It is given that function $f(x) = 2x^3 - 3x^2 - 36x + 7$
$\Rightarrow f'(x) = 6x^2 - 6x - 36$
$\Rightarrow f'(x) = 6(x^2 - x - 6)$
$\Rightarrow f'(x) = 6(x + 2) (x - 3)$
$\Rightarrow f'(x) = 6(x + 2)(x - 3)$
If $f'(x) = 0$, then we get,
$\Rightarrow x = -2, 3$
So, the points $x = -2$ and $x = 3$ divides the real line into two disjoint intervals, $(-\infty, -2),(-2,3) $ and $(3, \infty)$
So, in interval $(-\infty, -2) and (3, \infty)$
$f^\prime$(x) = 6(x + 2)(x - 3) > 0
But, in $(-2, 3)$, $f^\prime(x)<0$
Therefore, the given function 'f' is strictly increasing in interval $(-\infty, 2)and(3, \infty)$ while as strictly decreasing in $(-2, 3)$
$\Rightarrow f'(x) = 6x^2 - 6x - 36$
$\Rightarrow f'(x) = 6(x^2 - x - 6)$
$\Rightarrow f'(x) = 6(x + 2) (x - 3)$
$\Rightarrow f'(x) = 6(x + 2)(x - 3)$
If $f'(x) = 0$, then we get,
$\Rightarrow x = -2, 3$
So, the points $x = -2$ and $x = 3$ divides the real line into two disjoint intervals, $(-\infty, -2),(-2,3) $ and $(3, \infty)$
So, in interval $(-\infty, -2) and (3, \infty)$
$f^\prime$(x) = 6(x + 2)(x - 3) > 0
But, in $(-2, 3)$, $f^\prime(x)<0$
Therefore, the given function 'f' is strictly increasing in interval $(-\infty, 2)and(3, \infty)$ while as strictly decreasing in $(-2, 3)$
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