Let f(x) = √sinx. Use the tools of calculus to find the absolute minimum of f(x) on the interval [0,pi].

Question

Let f(x) = √sinx. Use the tools of calculus to find the absolute minimum of f(x) on the interval [0,pi].

Answer 1

For a continuous function on a closed interval its absolute max and min at that interval happens at either endpoint or at one of its critical numbers. To find critical numbers determine values of x that make the derivative equal to zero or make the derivative undefined.

The derivative of f(x) = √[sin(x)] or sin(x)^1/2:

f'(x) = 1/2[sin(x)]^-1/2cos(x) = cos(x)/2√sin(x).

The derivative equals zero when the numerator equals zero.

cos(x) = 0

x = cos^-1(0) = π/2 (the only value on the closed interval from 0 to π).

The first derivative is undefined when sin(x) = 0, x = 0 or π; these are also the endpoints of the closed interval.

Evaluate f(0, f(π/2), and f(π) to determine which value gives a minimum.

f(0) = √sin(0) = √0 = 0

f(π/2) = √(1) = 1

f(π) = √0 = 0

So, on the closed interval [0,π] the absolute min is 0 which happens at either endpoint.

Note that if the problem had asked for absolute max, the answer would be 1 happening at x = π/2.

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Answer 2

The absolute minimum of f(x)=sqrt{sin x}​ on [0,π] is 0, occurring at x=0 and x=π.