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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find both the maximum value and the minimum value of $3x^4 - 8x^3 + 12x^2 - 48x + 25$ on the interval $[0, 3].$

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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:
$f(x)=3x^4-8x^3+12x^2-48x+25$
$f'(x)=12x^3-24x^2+24x-48$
$\qquad=12(x^3-2x^2+2x-4)$
$\qquad=12[x^2(x-2)+2(x-2)]$
$\qquad=12(x^2+2)(x-2)$
For maxima and minima
$f'(x)=0$
$12(x^2+2)(x-2)=0$
$\Rightarrow x=2$
Step 2:
Now we find $f(x)$ at $x=0,2$ & 3
$f(0)=25$
$f(2)=3(2)^4-8(2)^3+12(2)^2-48(2)+25$
$\qquad=48-64+48-96+25$
$\qquad=-39$
$f(3)=3(3)^4-8(3)^3+12(3)^2-48(3)+25$
$\quad\;\;\;=243-216+108-144+25$
$\quad\;\;\;=16$
Hence at $x=0$ Maximum value=25
At $x=2$,Minimum value=-39

 

answered Aug 7, 2013 by sreemathi.v
edited Aug 17, 2013 by sharmaaparna1
 

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