What is the <strong>GREATEST</strong> value of the function?
Consider the following for the next two (02) items that follow: Let f(x)=\sqrt{2-x}+\sqrt{2+x}.
- A. \sqrt{3}
- B. \sqrt{6}
- C. \sqrt{8} ✓
- D. 4
Correct Answer: C. \sqrt{8}
Explanation
Differentiating the function gives f'(x) = \frac{-1}{2\sqrt{2-x}} + \frac{1}{2\sqrt{2+x}}. Setting f'(x) = 0 yields \sqrt{2-x} = \sqrt{2+x} \implies 2-x = 2+x \implies x = 0. The maximum value occurs at x = 0, giving f(0) = \sqrt{2} + \sqrt{2} = 2\sqrt{2} = \sqrt{8}. Checking the endpoints x = \pm 2 gives f(\pm 2) = \sqrt{4} = 2 = \sqrt{4}, which is smaller.
Related questions on Calculus
- Let z=[y] and y=[x]-x, where [.] is the greatest integer function. If x is <strong>NOT</strong> an integer but positive, then what i...
- If f(x)=4x+1 and g(x)=kx+2 such that fog(x)=gof(x), then what is the value of k?
- What is the <strong>MINIMUM</strong> value of the function f(x)=\log_{10}(x^{2}+2x+11)?
- What is \int(x^{x})^{2}(1+\ln x)\,dx equal to ?
- What is \int e^{x}\{1+\ln x+x\ln x\}\,dx equal to?