I need help ASAP For which function and interval can the extreme value theorem be applied?

f (x) = StartFraction x Over (x minus 6) squared EndFraction over the interval [3, 5]
f (x) = StartFraction negative 5 Over 2 (x minus 7) cubed EndFraction over the interval (6, 8)
f(x) = 4 tan(πx) over the interval [0, 1]
f(x) = 3ex + 1 over the interval (1, 3)

A) f(x)=x/(x-6)^2 over the interval [3, 5]

Step-by-step explanation:

got it right on edge :) good luck!

A

Step-by-step explanation:

I got it correct

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°