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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: $(iv)\: f (x) = (x-1)^2+3, x \in [– 3, 1] $

This is fourth part of multipart q5

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
$f(x)=(x-1)^2+3$
$f'(x)=2(x-1)$
For extreme values $f'(x)=0$
$f'(x)=2(x-1)$
$2(x-1)=0$
$(x-1)=0$
$x=1$
Step 2:
Now we find $f(x)$ at $x=1,-3$
$f(1)=(1-1)^2+3$
$\qquad=0+3=3$
Step 3:
$f(-3)=(-3-1)^2+3$
$\qquad\;\;=(-4)^2+3$
$\qquad\;\;=16+3$
$\qquad\;\;=19$
Step 4:
$\therefore$ Absolute maximum value =19 at $x=-3$
Absolute minimum value=3 at $x=1$
answered Aug 7, 2013 by sreemathi.v
 

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