Question

Determine whether the mean value theorem applies to the function on the interval

Answer 1

Given:

`H(x)=(4)/(x^2+4)`

• 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;

`f^(\prime)(c)=(f(b)-f(a))/(b-a)`

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).

`H^(\prime)(c)=((x^2+4).0-4(2c))/((c^2+4)^2)`
`=(-8c)/((c^2+4)^2)`

• Next, is to find H(3)

Substitute x=3 into the function H(x)

`H(3)=(4)/(3^2+4)=(4)/(9+4)=(4)/(13)`

• Similarly find H(-3)

`H(-3)=(4)/((-3)^2+4)=(4)/(9+4)=(4)/(13)`

Substitute the values we've gotten so far into;

`H^(\prime)(c)=(H(3)-H(-3))/(3-(-3))`
`(-8c)/((c^2+4)^2)=((4)/(13)-(4)/(13))/(9)`

• Evaluate

`(-8c)/((c^2+4)^2)=(0)/(9)`

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