Question

The maximum value of the function \(f(x)=x+\sqrt{1-x}\) on the interval [0,1] is, 

• A. \(\frac{1}{4}\)
• B. \(\frac{5}{4}\)
• C. \(\frac{3}{4}\)
• D. 1

Answer 1

The Correct Option is B

To find the maximum value of the function \(f(x)=x+\sqrt{1-x}\) on the interval \([0,1]\), we follow these steps:

1. Compute the derivative \(f'(x)\) to find critical points:
   \(f'(x)=1-\frac{1}{2\sqrt{1-x}}\).
2. Set the derivative to zero to find critical points:
   \(1-\frac{1}{2\sqrt{1-x}}=0\) leads to \(\frac{1}{2\sqrt{1-x}}=1\).
   Solving gives \(\sqrt{1-x}=\frac{1}{2}\).
   Squaring both sides, \(1-x=\frac{1}{4}\).
   Solve for \(x\) to get \(x=\frac{3}{4}\).
3. Evaluate \(f(x)\) at critical points and endpoints of the interval:
   \(f(0)=0+\sqrt{1}=1\), \(f(1)=1+\sqrt{0}=1\), \(f\left(\frac{3}{4}\right)=\frac{3}{4}+\sqrt{\frac{1}{4}}=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\).
4. Compare the values: \(f(0)=1\), \(f(1)=1\), \(f\left(\frac{3}{4}\right)=\frac{5}{4}\).
   The maximum value of \(f(x)\) on \([0,1]\) is \(\frac{5}{4}\).