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- By Mean value theorem $f'(c)=\large\frac{f(b)-f(a)}{b-a}$

Step 1:

$f(x)=x^2-1$ for $x\in [1,2]$

It is polynomial.Therefore it is continuous in the interval [1,2].

$f'(x)=2x$

$f(1)=1-1=0$

$f(2)=4-1=3$

$f'(c)=2c$

Step 2:

$f'(c)=\large\frac{f(b)-f(a)}{b-a}$

$2c=\large\frac{3-0}{2-1}$

$2c=\large\frac{3}{1}$

$2c=3$

$c=\large\frac{3}{2}$

$c=\large\frac{3}{2}$ which belongs to $(1,2)$

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