Question
Find the intervals in which the function $f$ given by $f(x)=2 x^3-3 x^2-36 x+7$ is decreasing.
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Answer
It is given that function $f(x)=2 x^3-3 x^2-36 x+7$
$\Rightarrow f^{\prime}(x)=6 x^2-6 x+36$
$\Rightarrow f^{\prime}(x)=6\left(x^2-x+6\right)$
$\Rightarrow f^{\prime}(x)=6(x+2)(x-3)$
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 $(-2, 3).$
$\Rightarrow f^{\prime}(x)=6 x^2-6 x+36$
$\Rightarrow f^{\prime}(x)=6\left(x^2-x+6\right)$
$\Rightarrow f^{\prime}(x)=6(x+2)(x-3)$
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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