To find the maximum value of the function [tex]\((x-1)^2 + x\)[/tex] within the interval [tex]\(0 < x < 1\)[/tex], let's go through the solution step by step:
1. Define the function:
[tex]\[
f(x) = (x - 1)^2 + x
\][/tex]
2. Find the critical points:
To find the critical points, we first need to take the derivative of the function [tex]\(f(x)\)[/tex].
[tex]\[
f'(x) = \frac{d}{dx}\left[(x-1)^2 + x\right]
\][/tex]
Using the chain rule and power rule:
[tex]\[
f'(x) = 2(x - 1) \cdot 1 + 1 = 2(x - 1) + 1 = 2x - 2 + 1 = 2x - 1
\][/tex]
3. Set the derivative equal to zero to find critical points:
[tex]\[
2x - 1 = 0 \implies x = \frac{1}{2}
\][/tex]
Since [tex]\(0 < x < 1\)[/tex], [tex]\(x = \frac{1}{2}\)[/tex] is within the interval.
4. Evaluate the function at the critical points and boundaries:
We need to evaluate the function [tex]\(f(x)\)[/tex] at [tex]\(x = \frac{1}{2}\)[/tex], and also at the boundaries [tex]\(x = 0\)[/tex] and [tex]\(x = 1\)[/tex].
- At [tex]\(x = \frac{1}{2}\)[/tex]:
[tex]\[
f\left(\frac{1}{2}\right) = \left(\frac{1}{2} - 1\right)^2 + \frac{1}{2} = \left(-\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}
\][/tex]
- At [tex]\(x = 0\)[/tex]:
[tex]\[
f(0) = (0 - 1)^2 + 0 = 1 + 0 = 1
\][/tex]
However, note the question specifies [tex]\(0 < x < 1\)[/tex], so we don't include this point for maximizing within the strict interval. Let's just formally note that [tex]\(x=0\)[/tex] would yield [tex]\(f(0)=1\)[/tex].
- At [tex]\(x = 1\)[/tex]:
[tex]\[
f(1) = (1 - 1)^2 + 1 = 0 + 1 = 1
\][/tex]
5. Compare the values to find the maximum:
[tex]\[
f\left(\frac{1}{2}\right) = \frac{3}{4}
\][/tex]
[tex]\[
f(0) = 1
\][/tex]
[tex]\[
f(1) = 1
\][/tex]
Thus, the maximum value of the function within the interval [tex]\(0 < x < 1\)[/tex] is [tex]\(\boxed{1}\)[/tex].
Hence the correct answer is:
c. 1
