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Verify Mean Value Theorem, if \( f (x) = x^3 - 5x^2 - 3x \) in the interval \( [a,b]\), where \( a = 1\) and \( b = 3\). Find all \( c \in (1,3) \) for which \( f'(c) = 0\).

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Toolbox:
  • $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$f(x)=x^3-5x^2-3x$
It is a polynomial.Therefore it is continuous in the interval $[1,3]$ and derivative in the interval (1,3).
$f(x)=x^3-5x^2-3x$
$f'(x)=3x^2-10x-3$
 
Step 2:
Let $a=1 ,b=3$
$f'(c)=3c^2-10c-3$
$f(1)=(1)^3-5(1)^2-3(1)$
$\qquad=1-5-3$
$\qquad=-7$
$f(3)=(3)^3-5(3)^2-3(3)$
$\qquad=27-45-9$
$\qquad=-27$
Step 3:
By Mean value theorem $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
$3c^2-10c-3=\large\frac{-27-(-7)}{3-1}$
$\qquad\qquad\quad\;\;=\large\frac{-20}{2}$
$\qquad\qquad\quad\;\;=-10$
$3c^2-10c-3=-10$
$3c^2-10c-3+10=0$
$3c^2-10c+7=0$
$3c^2-7c-3c+7=0$
$3c^2-3c-7c+7=0$
$3c(c-1)-7(c-1)=0$
$(3c-7)(c-1)=0$
$c=\large\frac{7}{3}$ or $c=1$
Now $c=\large\frac{7}{3}$$\in (1,3)$
$c=\large\frac{7}{3}$
Step 4:
$f'(c)=0\Rightarrow 3c^2-10c-3=0$
$c=\large\frac{10\pm\sqrt{100+36}}{6}$
$c=\large\frac{10\pm\sqrt{136}}{6}$
$\;=\large\frac{10\pm2\sqrt{34}}{6}$
$\;=\large\frac{5\pm\sqrt{34}}{3}$
$\;=3.61-0.28$
None of these values $\in (1,3)$
answered May 13, 2013 by sreemathi.v
 

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