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

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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 $[-3,-1]$ we find the values of $f(x)$ at $x=-3,-2,-1$
$f(-3)=2(-3)^3-24(-3)+107$
$\qquad\;=-54+72+107$
$\qquad\;=-125$
$f(-2)=2(-2)^3-24(-2)+107$
$\qquad\;=-16+48+107$
$\qquad\;=139$
$f(-1)=2(-1)^3-24(-1)+107$
$\qquad\;=129$
Hence maximum $f(x)=139$ at $x=-2$
answered Aug 19, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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