Given:
[tex]H(x)=\frac{4}{x^2+4}[/tex]• First Let's go through the, mean value theorem.
If a function f is continuous on the closed interval [a, b] and is differenciable on the open interval (a, b), the there is a number c in (a, b) such that;
[tex]f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}[/tex]The function given satisfies all this conditions.
a=-3 and b= 3
• Let's ,differentiate, the function given using quotient rule to get f'(c).
[tex]H^{\prime}(c)=\frac{(x^2+4).0-4(2c)}{(c^2+4)^2}[/tex][tex]=\frac{-8c}{(c^2+4)^2}[/tex]• Next, is to find H(3)
Substitute x=3 into the function H(x)
[tex]H(3)=\frac{4}{3^2+4}=\frac{4}{9+4}=\frac{4}{13}[/tex]• Similarly find H(-3)
[tex]H(-3)=\frac{4}{(-3)^2+4}=\frac{4}{9+4}=\frac{4}{13}[/tex]Substitute the values we've gotten so far into;
[tex]H^{\prime}(c)=\frac{H(3)-H(-3)}{3-(-3)}[/tex][tex]\frac{-8c}{(c^2+4)^2}=\frac{\frac{4}{13}-\frac{4}{13}}{9}[/tex]• Evaluate
[tex]\frac{-8c}{(c^2+4)^2}=\frac{0}{9}[/tex]From the above;
-8c = 0 ⇒ c =0
(c² + 4)² = 9
Take the root of both-side
c² + 4 = 3
c² = 3 - 4
c² = -1
c=±√-1
c = ± i
But from the definition c is in [-3, 3]
Hence, c = 0