15.2k views
4 votes
Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. f(x) = 5x + 4 h(x) = x2 + 8x + 24

x = -3
x = 3
x = 4
x = 5

User LdM
by
8.2k points

2 Answers

0 votes

The correct answer is x = 5.

Trust me, I just did the quiz.

User Artyom Shalkhakov
by
8.1k points
2 votes
The exponential function is
f(x)= 5^(x) +4
The polynomial function is
h(x)= x^(2) +8x+24

We want a value of x as such that
f(x)\ \textgreater \ h(x), so


5^(x)+4 >
x^(2) +8x+24

We are provided with four possible value of x. By process of trial and error, we will narrow down two values of x when the statement is changing from false to true

first, we try
x=-3 by substituting into
f(x) and
h(x)


f(-3)=5^(-3)+4 = 4.008 and
h(-3)= (-3)^(2)+8(-3)+24=9
so
f(x)>
h(x) which is not the inequality wanted

We try
x=3 by substituting into
f(x) and
h(x)

f(3)= 5^(3)+4=129 and
h(3)= (3)^(2)+8(3)+24=57,
so
f(x) >
h(x) as wanted

We know that any values of
x greater than three will give
f(x) greater than
h(x), so we narrow down to two values that make the statement changes from false to true.

There is one value between -3 and three that makes
f(x) greater than
h(x). From here we use the method of trial and error.

We can start with the mid value between -3 and three which is 0

f(0)= 5^(0)+4=5

h(0)= 0^(2)+8(0)+24=24

f(x) is less than
h(x) which isn't the inequality we want so here we narrow down the value of x must be between 0 and 3

We can try mid-value between 0 and three which is 1.5

f(1.5)= 5^(1.5)+4=15.18

h(1.5)= (1.5)^(2)+8(1.5)+24=38.25

f(x) is still less than
h(x), so we now narrow down the value of x between 1.5 and 3

We try again using the mid-value between 1.5 and three which is 2.25

f(2.25)= 5^(2.25)+4=41.38

h(2.25)= 2.25^(2)+8(2.25)+24=47.0625
We still have
f(x) less than
h(x) so we can narrow down further the value of x between 2.25 and 3 (which is quite a short interval already)

We can keep trying by choosing a value of x says x=2.6

f(2.6)= 5^(2.6)+4=69.66

h(2.6)= 2.6^(2)+8(2.6)+24=51.56
We have
f(x) greater then
h(x)

The narrowing down process continues where we now have the interval of x between 2.25 and 2.6. Keeping to one decimal place we can find the
value of x=2.4 is the approximate value where
f(x) is greater than
h(x).

Answer: x=2.4
User Alexandre Mazel
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories