The extreme value theorem can be applied on an interval if the function is continuous in the entire interval. Testing the continuity for each function, we get that the correct option is the third option.
Continuity:
A function f is continuous at an interval if all points in the interval are in the domain of the function, and, for each point of [tex]x^{\ast}[/tex], the limit exists and:
[tex]\lim_{x \rightarrow x^{\ast}} = f(x^{\ast})[/tex]
First function:
At [tex]x = 4[/tex], the denominator is 0, and so, the extreme value theorem cannot be applied.
Second function:
At [tex]x = 5[/tex], the denominator is 0, and so, the extreme value theorem cannot be applied.
Third function:
The only point there can be a discontinuity is at [tex]x = 4[/tex], where the definition of the function changes. First we have to find the lateral limits, and if they are equal, the limits exist:
To the left(-), it is less than 4, so we take the definition for x < 4.
To the right(+), it is more than 4, so we take the definition for x >= 4.
[tex]\lim_{x \rightarrow 4^{-}} h(x) = \lim_{x \rightarrow 4} \frac{9x}{10-x} = \frac{9*4}{10-4} = \frac{36}{6} = 6[/tex]
[tex]\lim_{x \rightarrow 4^{+}} h(x) = \lim_{x \rightarrow 4} x + 2 = 4 + 2 = 6[/tex]
The lateral limits are equal, so the limit exists, and it's value is 6.
The definition at [tex]x = 4[/tex] is [tex]h(x) = x + 2[/tex], so [tex]h(4) = 4 + 2 = 6[/tex].
Since [tex]\lim_{x \rightarrow 4} = h(4)[/tex], the function is continuous over the entire interval, and this is the correct answer.
Fourth function:
There can be a discontinuity at [tex]x = 5[/tex], so we test the limits:
[tex]\lim_{x \rightarrow 5^{-}} h(x) = \lim_{x \rightarrow 5} -x = -5[/tex]
[tex]\lim_{x \rightarrow 5^{+}} h(x) = \lim_{x \rightarrow 5} x^2 - 20 = 5^2 - 20 = 25 - 20 = 5[/tex]
Different limits, so the limit does not exist and the function is not continuous at [tex]x = 5[/tex], and the extreme value theorem cannot be applied.
For more on the extreme value theorem, you can check https://brainly.com/question/15585098