Liam is using sequences to compare the growth rates of h(x)=1.2x

Question

and j(x)=1.2^x
. Which statement correctly describes how Liam should do this and what he will observe?(1 point)
Responses

Liam should look at where one sequence has terms greater than the terms in the other sequence. The growth rate of j(x)=1.2^x
is only greater than the growth rate of h(x)=1.2x
when its terms are greater.
Liam should look at where one sequence has terms greater than the terms in the other sequence. The growth rate of j times x is equal to 1 point 2 to the x th power is only greater than the growth rate of h of x is equal to 1 point 2 x when its terms are greater.

Liam should look at where one sequence has terms greater than the terms in the other sequence. The growth rate of h(x)=1.2x
is only greater than the growth rate of j(x)=1.2^x
when its terms are greater.
Liam should look at where one sequence has terms greater than the terms in the other sequence. The growth rate of h of x is equal to 1 point 2 x is only greater than the growth rate of j times x is equal to 1 point 2 to the x th power when its terms are greater.

Liam should compare the rates of change of the terms in both sequences. The growth rate of j(x)=1.2^x
will quickly surpass the growth rate of h(x)=1.2x
Liam should compare the rates of change of the terms in both sequences. The growth rate of j times x is equal to 1 point 2 to the x th power will quickly surpass the growth rate of h of x is equal to 1 point 2 x

Liam should compare the rates of change of the terms in both sequences. The growth rate of h(x)=1.2x
will quickly surpass the growth rate of j(x)=1.2^x
Liam should compare the rates of change of the terms in both sequences. The growth rate of h of x is equal to 1 point 2 x will quickly surpass the growth rate of j times x is equal to 1 point 2 to the x th power

Answer 1

Liam should compare the rates of change of the terms in both sequences. The growth rate of $j(x)={1.2}^x$ will quickly surpass the growth rate of $h(x)=1.2x$.

This statement is accurate because exponential functions, like ${1.2}^x$, grow faster than linear functions, like $1.2x$, as $x$ increases. Therefore, while at smaller values of $x$, $h(x)$ may be larger, $j(x)$ will eventually exceed $h(x)$ as $x$ becomes large.