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# Find the intervals in which the function f given by $f (x) = 2x^3 – 3x^2 – 36 x + 7$ is (a) strictly increasing and (b) strictly decreasing ?

This question has appeared in model paper 2012

$\begin{array}{1 1} (A)\; \text{Strictly increasing in }(-2,3) \text{and strictly decreasing in} (-\infty,-2)\; and \;(3,\infty) \\ (B)\; \text{Strictly increasing in} (-\infty,-2) \;and\; (3,\infty)\; \text{and strictly decreasing in} (-2,3) \\ (C)\; \text{Strictly increasing in } (-\infty,-2)\; and\; (-3,2) \text{ and strictly decreasing in} (-2,3) \\ (D) \text{None of the above. } \end{array}$

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## 1 Answer

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Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Given : $f(x)=2x^3-3x^2-36x+7$
Differentiating w.r.t $x$ we get,
$f'(x)=6x^2-6x-36$
$\qquad\;=6(x^2-x-6)$
On factorizing we get,
$\qquad\;=6(x-3)(x+2)$
$\Rightarrow f'(x)=0$ at $x=3$ and $x=-2$
Step 2:
Hence the real number line is divided into three intervals.
(i.e)$(-\infty,-2),(-2,3),(3,\infty)$
Consider the interval $(-\infty,-2)$ and $(3,\infty)$
We can see that each factor $(x-3)$ and $(x+2)$ is negative and hence their product is positive.
$f'(x)(-3)(x+2)>0$
$\Rightarrow f'(x)>0$.
Hence it is strictly increasing in the interval $(-\infty,-2)$ and $(3,\infty)$
Step 3:
Consider the interval $(-2,3)$
Here $(x+2)$ is positive and $(x-3)$ negative.Hence this product is negative .
Therefore $f'(x)<0$
Therefore $f(x)$ is strictly decreasing in the interval $(-2,3)$
answered Jul 9, 2013

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