The maximum value of the function \(f(x)=x+\sqrt{1-x}\) in the interval [0, 1] is,
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{5}{4}\)
On a closed interval, the absolute maximum is the largest of the values at the critical points and at the end points.
\( \displaystyle f'(x) = 1 - \frac{1}{2\sqrt{1-x}} \)
Setting \(f'(x) = 0\):
\(2\sqrt{1-x} = 1\)
\( \displaystyle 1 - x = \frac{1}{4} \)
\( \displaystyle x = \frac{3}{4} \)
Values:
\(f(0) = 0 + 1 = 1\)
\( \displaystyle f\left(\frac{3}{4}\right) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} \)
\(f(1) = 1 + 0 = 1\)
The maximum value is \(\frac{5}{4}\).
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{4}\) — Incorrect. \(\frac{1}{4}\) is the value of 1 − x at the critical point, not a value of f.
Option (3) \(\displaystyle\frac{3}{4}\) — Incorrect. \(\frac{3}{4}\) is the point x where the maximum occurs, not the maximum value.
Option (4) 1 — Incorrect. 1 is the value at the end points, and \(f\left(\frac{3}{4}\right) = \frac{5}{4}\) is larger.