Math I really need help

Question

The functions f(x) = 2x² + 2 and g(x) = 2^x+1 + 2 are graphed on the coordinate plane as shown.

For what values of x is g(x) > f(x)?

1. A. x > 4

1. B. 0 < x < 2 and x > 4

1. C. x > 2

1. D. 2 < x < 4

Answer 1

set g(x) = f(x) to find where they intersect

2^(x+1) + 2 = 2x^2 + 2

subtract 2 from both sides

2^(x+1) = 2x^2

divide both sides by 2

2^x = x^2.

somewhat "obviously" x=2 is one solution. It's quadratic so it has a 2nd solution

2^2 =2^2 = 4

also

2^4 = 4^2 = 16, x=4 is the other solution

3^2>2^3, 9>8

1^2<2^1, 1<2

5^2<2^5, 25<32

g(x)> f(x) between x=2 and x=4

f(x)>g(x) for x<2 and x>4

f(x)=g(x) when x=2 or x=4

D is the correct answer 2<x<4

Or you could figure this all out just looking at the graph

Answer 2

If you plug in any number that is greater than 4, than g(x)>f(X). If you also plug in the number one, which is between 0 and 2 than g(x)>f(x)

For example if you plug in the number five for f(x) the answer is 52 because (2 multiplied by 5 to the 2nd power plus 2 equals 52).

If you plug in the number 5 for g(x), the answer is 66.

When you plug in the number one, f(x) is equal to 4 and g(x) is equal to 6.

Therefore, the answer is B. The answer is not C and D because if you plug in 3, then f(x) will be greater than g(x).

Answer 3

If you look at the graph, you can see that the graphs of f(x) and g(x) cross at two points, at x = 2 and again at x = 4. So you need to look at the intervals between x= 0 and x = 2, from x= 2 to x= 4, and again for x>4.

1. Between x=0 and x = 2 (0 < x < 2), the g(x) graph is above the f(x) graph, so g(x) is greater than f(x) (g(x) > f(x)).
2. Between x = 2 and x = 4 (2 < x< 4), the f(x) graph is on top, so f(x) is greater than g(x) (f(x) > g(x)).
3. From x=4 onward (x>4), g(x) is on top again, so its greater (g(x) > f(x)).