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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the intervals in which the following functions are strictly increasing or decreasing: $(e) \;(x+1)^3(x-3)^3$

This is (e) part of the multi-part question q6

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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 : $f(x)=(x+1)^3(x-3)^3$
Differentiating w.r.t $x$ by applying product rule
$f'(x)=(x+1)^3.3(x-3)^2+(x-3)^3.3(x+1)^2$
$\qquad=3(x+1)^3(x-3)^2+3(x-3)^3.(x+1)^2$
$\qquad=3(x+1)^2(x-3)^2[x-3+x+1]$
$\qquad=6(x+1)^2(x-3)^2(x-1)$
Step 2:
If $f(x)$ is increasing :
$f'(x) >0$
$\Rightarrow 6(x+1)^2(x-3)^2(x-1) >0$
But $6(x+1)^2(x-2)^2 >0$
Hence $(x-1) >0$
$\Rightarrow x >1$
But $f'(x)=0$ at $x=3$
$\Rightarrow f(x)$ is increasing in $(1,3)$ and $(3,0)$
Therefore $f(x)$ is increasing on $(1,\infty)$
Step 3:
When $f(x)$ is decreasing :
$f'(x) <0$
$\Rightarrow 6(x+1)^2(x-3)^2(x-1)<0$
(i.e) $(x-1) <0$
$\Rightarrow x < 0$
But $f'(x)=0$ at $x=-1$
$\Rightarrow f(x) $ is strictly decreasing in $(-\infty,1),(-1,1)$
answered Jul 9, 2013 by sreemathi.v
 

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