Discussion Forum

The value  of c from the lagrange's mean value theorm for which $f(x)=\sqrt{25-x^{2}}$ in [1,5] is 

Answer:

Explanation:

It is clear that f(x) has a definite  and unique value for each x$\in [1,5]$

Thus, for every point in the interval [1,5] the value of f(x) exists

 So, f(x)  is continuous in the interval [1,5]

Also, $f'(x)=\frac{-x}{\sqrt{25-x^{2}}}$, which clearly exists  for all x in an open interval (1,5)

So,  f'(x) is differentiable in [1,5]

So, there must be a value $c \in [1,5]$ such that  

$f'(c)= \frac{f(5)-f(1)}{5-1}=\frac{0-\sqrt{24}}{4}$

$=\frac{0-2\sqrt{6}}{4}=\frac{-\sqrt{6}}{2}$

 but f'(c)$=\frac{-c}{\sqrt{25-c^{2}}}$

 $\Rightarrow$    $\frac{-c}{\sqrt{25-c^{2}}}=-\frac{\sqrt{6}}{2}$

 $\Rightarrow$    $4c^{2}=6(25-c^{2})$

 $\Rightarrow$  $4c^{2}=150- 6c^{2} \Rightarrow 10c^{2}=150$

 $\Rightarrow$   $c^{2}=15 \Rightarrow c= \pm \sqrt{15}$

  $\therefore$     $c= \sqrt{15} \in [1,5]$