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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: $\: f (x) = x^3, x \in [– 2, 2] $

$\begin{array}{1 1} Max=8 \;Min=-7 \\ Max=8 \;Min=-8 \\ Max=6 \;Min=-8 \\ Max=8 \;Min=8 \end{array} $

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • For Maxima and Minima $f'(x)=0$
Step 1:
We have $f(x)=x^3$ in $[-2,2]$
Differenting with respect to x we get
$f'(x)=3x^2$
For maxima and minima
$f'(x)=0$
At $x=0$
$f(0)=0$
Step 2:
Hence the given interval is $(-2,2)$
$f(-2)=(-2)^3$
$\quad\quad\;\;=-8$
$f(0)=0^3=0$
$f(2)=(2)^3=8$
Hence the absolute maximum value of $f(x)$ is 8 which is attained at $x=2$ and absolute minimum value of $f(x)=-8$ which is attained at $x=-2$
answered Aug 7, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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