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# Find the intervals in which the following function is (a) increasing (b) decreasing : $f(x) = x^3-12x^2+36x+17$

Can you answer this question?

Toolbox:
• If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
• If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$
Step 1:
$f(x)=x^3-12x^2+36x+17$
$f'(x)=3x^2-24x+36$
$f'(x)=0$
$\Rightarrow 3x^2-24x+36=0$
(i.e)$3(x^2-8x+12)=0$
On factorizing we get,
$3(x-6)(x-2)=0$
Step 2:
Hence the point $2,6$ divides the real number line into three intervals :
$(-\infty,2),(2,6),(6,\infty)$
In the interval $(-\infty,2)$ sign of f'(x) is (-)(-)>0$\Rightarrow$ the nature of the function $f$ is strictly increasing.
In the interval $(2,6)$ sign of f'(x) is (-)(+)<0$\Rightarrow$ the nature of the function $f$ is strictly decreasing.
In the interval $(6,\infty)$ sign of f'(x) is (+)(+)>0$\Rightarrow$ the nature of the function $f$ is strictly increasing.
answered Oct 2, 2013