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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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  • $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
It is a polynomial.Therefore it is continuous in the interval $[1,3]$ and derivative in the interval (1,3).
Step 2:
Let $a=1 ,b=3$
Step 3:
By Mean value theorem $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
$c=\large\frac{7}{3}$ or $c=1$
Now $c=\large\frac{7}{3}$$\in (1,3)$
Step 4:
$f'(c)=0\Rightarrow 3c^2-10c-3=0$
None of these values $\in (1,3)$
answered May 13, 2013 by sreemathi.v

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