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