Question

help with this one, please

Answer 1

At x = 3 the exponential function h(x) begin to exceed the linear function f(x).

Solution:

Given data:

$f(x)=2x+25$

$h(x)=5^x$

To find at what value of x in h(x) begin to exceed the linear function f(x).

Substitute x = 1 in f(x) and h(x).

$f(1)=2(1)+25=27$

$h(1)=5^1=5$

27 > 5

f(x) is not exceed h(x), so proceed further.

Substitute x = 2 in f(x) and h(x).

$f(2)=2(2)+25=29$

$h(2)=5^2=25$

29 > 25

f(x) is not exceed h(x), so proceed further.

Substitute x = 3 in f(x) and h(x).

$f(3)=2(3)+25=31$

$h(3)=5^3=125$

31 < 125

f(x) is exceed h(x), the process stops here.

Here the value of x is 3.

Hence at x = 3 the exponential function h(x) begin to exceed the linear function f(x).