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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
+1 vote

Find the maximum value of \(2x^3 – 24x + 107\) in the interval \([1, 3]\).

This is separated into two parts.

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • Maxima & Minima=$f'(x)=0$
Step 1:
$f(x)=2x^3-24x+107$
Differentiating with respect to $x$
$f'(x)=6x^2-24$
For maxima & minima
$f'(x)=0$
$6x^2-24=0$
$6x^2=24$
$x^2=\large\frac{24}{6}$
$x^2=4$
$x=\pm 2$
Step 2:
For the interval $[1,3]$ we find the values of $f(x)$ at $x=1,2,3$
$f(1)=2(1)^3-24\times 1+107$
$\qquad=2-24+107$
$\qquad=85$
$f(2)=2(2)^3-24\times 2+107$
$\qquad=16-48+107$
$\qquad=75$
$f(3)=2(3)^3-24\times +107$
$\qquad=32-72+107$
$\qquad=89$
Hence maximum $f(x)=89$ at $x=3$.
answered Aug 19, 2013 by sreemathi.v
edited Aug 30, 2013 by sharmaaparna1
 

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