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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: $(iii)\: f (x) =4x-\large\frac{x^2}{2}$$, x \in [– 2, \large\frac{9}{2}] $

This is third part of multipart q5

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}(x^n)$$=nx^{n-1}$
Step 1:
On differentiating with respect to x we get
$f'(x)=4-\large\frac{1}{2}$$.2x$
$\qquad=4-\large\frac{2x}{2}$
$\qquad=4-x$
For extreme values $f'(x)=0$
$4-x=0$
$x=4$
Step 2:
Now we find the values of $f(x)$ at $x=-2,4,\large\frac{9}{2}$
$f(-2)=4(-2)-\large\frac{(-2)^2}{2}$
$\qquad=-8-\large\frac{4}{2}$
$\qquad=\large\frac{-16-4}{2}$
$\qquad=\large\frac{-20}{2}$
$\qquad=-10$
Step 3:
$f(4)=4(4)-\large\frac{(4)^2}{2}$
$\qquad=16-\large\frac{16}{2}$
$\qquad=\large\frac{32-16}{2}$$=8$
Step 4:
$f(\large\frac{9}{2})=$$4\times \large\frac{9}{2}-\large\frac{1}{2}$$\times \large \frac{81}{4}$
$\qquad=18-\large\frac{81}{8}$
$\qquad=\large\frac{144-81}{8}$
$\qquad=\large\frac{63}{8}$
$\qquad=7.875$
Step 5:
At $x=4$,absolute maximum value$=8$
At $x=2$, absolute minimum value $=-10$
answered Aug 7, 2013 by sreemathi.v
edited Aug 30, 2013 by sharmaaparna1
 

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