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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the points at which the function \(f\) given by \(f (x) = (x – 2)^4 (x + 1)^3\) has a (i) local maxima, (ii) local minima and the (iii) point of inflexion.

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • For maxima & minima $f'(x)=0$
Step 1:
Given:
Let $f(x)=(x-2)^4.(x+1)^3$
Differentiating with respect to x we get
$f'(x)=(x-2)^43(x+1)^2+(x+1)^3.4(x-2)^3$
$\qquad=(x-2)^3.(x+1)^2[3(x-2)+4(x+1)]$
$\qquad=(x-2)^3.(x+1)^2(7x-2)$
$\qquad=(x-2)^3.(x+1)^2[3(x-2)+4(x+1)]$
$\qquad=7(x-2)^3(x+1)^2(x-\large\frac{2}{7})$
For maxima and minima,$f'(x)=0$
$\Rightarrow 7(x-2)^3(x+1)^2(x-\large\frac{2}{7})$$=0$
$\Rightarrow x=2,-1,\large\frac{2}{7}$
Step 2:
At $x=2$
When $x$ is slightly $<2$
$f'(x)=(-)(+)(+)=-ve$
When $x$ is slightly $>2$
$f'(x)=(+)(+)(+)=+ve$
$f'(x)$ changes the sign from -ve to +ve while crossing the point $x=2$
$\Rightarrow f(x)$ is minimum at $x=2$
Step 3:
At $x=-1$
When $x$ is slightly $<-1$
$f'(x)=(-)(+)(-)=+ve$
When $x$ is slightly $>-1$
$f'(x)=(-)(+)(-)=+ve$
$\Rightarrow f'(x)$ does not changes the sign while passing through -1
Hence x = -1 is a point of inflexion.
Step 4:
Thus at $x=\large\frac{2}{7}$$=0.28$
When $x$ is slightly $<\large\frac{2}{7}$
$f'(x)=(-)(+)(-)=+ve$
When $x$ is slightly $>\large\frac{2}{7}$
$f'(x)=(-)(+)(+)=-ve$
While crossing the point $x=\large\frac{2}{7}$
$f(x)$ is maximum at $x=\large\frac{2}{7}$
answered Aug 12, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

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