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
9514 1404 393
Answer:
"complete the square" to put in vertex form
Step-by-step explanation:
It may be helpful to consider the square of a binomial:
(x +a)² = x² +2ax +a²
The expression x² +x +1 is in the standard form of the expression on the right above. Comparing the coefficients of x, we see ...
2a = 1
a = 1/2
That means we can write ...
(x +1/2)² = x² +x +1/4
But we need x² +x +1, so we need to add 3/4 to the binomial square in order to make the expressions equal:

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Another way to consider this is ...
x² +bx +c
= x² +2(b/2)x +(b/2)² +c -(b/2)² . . . . . . rewrite bx, add and subtract (b/2)²*
= (x +b/2)² +(c -(b/2)²)
for b=1, c=1, this becomes ...
x² +x +1 = (x +1/2)² +(1 -(1/2)²)
= (x +1/2)² +3/4
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* This process, "rewrite bx, add and subtract (b/2)²," is called "completing the square"—especially when written as (x-h)² +k, a parabola with vertex (h, k).