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 correct answer is x = 5.

Trust me, I just did the quiz.

Answer 2

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