Answer:
{
x
∣
x
>
1
} or x
∈
(
1
,
∞
)
Step-by-step explanation:
2
x
+
3
>
7
⇒
2
x
>
4
(subtract 3 from both sides)
2
x
+
3
>
7
⇒
x
>
2
(divide both sides by 2)
2
x
+
9
>
11
⇒ 2
x
>
2
(subtract 9 from both sides)
2
x
+
9
>
11
⇒
x
>
1
(divide both sides by 2)
if an x
-value is greater than 2, it will automatically be greater than 1. Thus, the solution set for 2
x+
3
>
7 is a subset of the one for 2
x
+
9
>11
.That means, all we need to do here is list the solution set for 2
x
+
9
>
11
, and we're done.
The solution set we need is simply "all x such that x is greater than 1", or {
x
∣
x
>
1
}
or
x
∈
(
1
,
∞
)
For the purpose, it is convenient to put the equation into intercept form:
x/a + y/b = 1
Here, we can divide by 2 and subtract the x-term to get
x/(-1) + y/2 = 1
Then we can see that the x-intercept is (-1, 0) and the y-intercept is (0, 2).
Substitute g(x) where x is in the definition of f(x).
f(g(x)) = -3(g(x))^2 - 1
f(g(x)) = -3(-5x)^2 - 1
= -3(25x^2) - 1
f(g(x)) = -75x^2 - 1