Question

99PTS!!!!!!!

Answer 1

Answer:

Option A.

Step-by-step explanation:

The given function is

$h(x)=3^x-2$        ... (1)

It is an exponential function.

The graph of second exponential function passes through the points (0,4) and (1,5), and it shifted 3 units up.

$g(x)=a(b)^x+3$

Substitute x=0 and g(x)=4 in the above function.

$4=a(b)^0+3\Rightarrow a=1$

Substitute a=1, x=1 and g(x)=5 in the above function.

$5=1(b)^1+3\Rightarrow b=2$

The second function is

$g(x)=2^x+3$             .... (2)

At x=0,

$h(0)=3^(0)-2=-1$

$g(0)=2^(0)+3=4$

So, at initial stage g(x)>h(x).

On solving (1) and (2) we get

$x=2,g(x)=h(x)=7$

It means at x=2 the function h(x) is equal to g(x) and after that h(x)>g(x).

g(x) ≥ h(x) on the interval –2 ≤ x ≤ 2

Therefore, the correct option is A.

Answer 2

h(x) = 3^x – 2 will be negative  when x is less than 0

h(2) =7

g(2)=7

This is the point they are equal

g(x)> h(x) until2

g(x)>=h(x) -2<=x<=2

Choice A