Question

PLEASE HELP DUE TODAY!!!!!!!

Consider the functions g(x) = 2x + 1 and h(x) = 2x + 2 for the domain 0 < x < 5


a. Without evaluating or graphing the functions, how do the ranges compare?


b. graph the 2 functions and describe each range over the given interval

Answer 1

• a. The range of h(x) over the given domain will have endpoints 1 unit greater.

• b. See the graph in the explanation. The range of g(x) over the given domain is (1, 11) and of h(x) is (2,12)

Explanation

a. Given the functions:

g(x) = 2x+1 h(x) = 2x + 2

for the domain 0 < x < 5.

These are two parallel lines where h(x) is 1 unit above g(x) Then, the range of h(x) over the given domain will have endpoints 1 unit greater.

b. Evaluating g(x) when x = 0 :

g(0) = 2(0) + 1 g(0) = 0 + 1 g(0) = 1

Evaluating g(x) when x = 5 :

g(5) = 2(5) + 1

g(5) = 10 + 1

g(5) = 11

Then, we can graph g(x) by connecting the points (0, 1) and (5, 11).

Evaluating h(x) when x = 0 :

h(0) = 2(0) + 2 h(0) = 0 + 2 h(0) = 2

Evaluating h(x) when x = 5 :

h(5) = 2(5) + 2

h(5) = 10 + 2

h(5) = 12

Then, we can graph h(x) by connecting the points (0, 2) and (5, 12).

See Graph

From the graph, the range of g(x) over the given domain is (1, 11) and of h(x) is (2,12)

Answer 2

a. The ranges of g(x) and h(x) differ by a constant value of 1, since h(x) is simply g(x) shifted upward by 1 unit. Therefore, the range unit.

b.

Graph of g(x) = 2x + 1 and h(x) = 2x + 2 over the interval 0 < x < 5:

![Graph of g(x) over the interval 0 < x < 5 is (1, 11), while the range of h(x) is (2, 12). This confirms that the range of h(x) is equal to the range of g(x) shifted upward by 1 unit.