Plug x = 0 into the function
f(x) = x^3 + 2x - 1
f(0) = 0^3 + 2(0) - 1
f(0) = -1
Note how the result is negative. The actual number itself doesn't matter. All we care about is the sign of the result.
Repeat for x = 1
f(x) = x^3 + 2x - 1
f(1) = 1^3 + 2(1) - 1
f(1) = 2
This result is positive.
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We found that f(0) = -1 and f(1) = 2. The first output -1 is negative while the second output 2 is positive. Going from negative to positive means that, at some point, we will hit y = 0. We might have multiple instances of this happening, or just one. We don't know for sure. The only thing we do know is that there is at least one root in this interval.
To actually find this root, you'll need to use a graphing calculator because the root is some complicated decimal value. Using a graphing calculator, you should find the root to be approximately 0.4533976515
The intermediate value theorem is used to determine if a value of the function exist in a given interval.
The function [tex]f(x) = x^3 + 2x - 1[/tex] has 0 on interval [tex][0,1][/tex]
Given
[tex]f(x) = x^3 + 2x - 1[/tex]
Calculate f(0)
[tex]f(x) = x^3 + 2x - 1[/tex]
[tex]f(0) =0^3 + 2 \times 0 - 1[/tex]
[tex]f(0) =-1[/tex]
Calculate f(1)
[tex]f(x) = x^3 + 2x - 1[/tex]
[tex]f(1) = 1^3 + 2 \times 1 - 1[/tex]
[tex]f(1) = 2[/tex]
So, we have:
[tex]f(0) =-1[/tex] and [tex]f(1) = 2[/tex]
By comparison:
[tex]f(0) < 0 < f(1)[/tex]
This means that:
The function has 0 on its interval.
Read more about intermediate value theorem at:
https://brainly.com/question/11001781
