Question

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

Answer 1

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

Answer 2

The correct answer is x = 5.

Trust me, I just did the quiz.