Question

Which statement correctly compares the two functions on the interval [-1,2]?

Answer 1

Answer:

Option A.

Step-by-step explanation:

Function f:

For x between -1 and 2, the values of f(x) increase, which means that f(x) is increasing.

Function g:

$g(x) = -18(\frac{1}{3})^x + 2$

Between -1 and 2:

$g(-1) = -18(\frac{1}{3})^{-1} + 2 = -18*3 + 2 = -54 + 2 = -52$

$g(0) = -18(\frac{1}{3})^{0} + 2 = -18*1 + 2 = -18 + 2 = -16$

$g(1) = -18(\frac{1}{3})^{1} + 2 = -18*\frac{1}{3} + 2 = -6 + 2 = -4$

$g(2) = -18(\frac{1}{3})^{2} + 2 = -18*\frac{1}{9} + 2 = -2 + 2 = 0$

Both are increasing.

However, g starts with a lower value, and finishes with a higher value, which means that function g increases at a faster average rate, and the correct answer is given by option A.