Question

Which of the following points are extrema for f(x) = sin(x) ?

• A. x = π/2
• B. x = 2π
• C. x = $\frac{3\pi }{2}$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A

x = π/2

C

x = $\frac{3\pi }{2}$

We’ll apply second derivative test to find out extremas. According to second derivative test, f’(x) = 0 at an extrema. If f”(x) is positive, it is a point of minima and if it is negative it is a point of maxima.

f(x) = sin(x)

f’(x) = cos(x)

1. f’(π/2) = 0

f”(x) = - sin(x)

f”(π/2) = -1

Since, f”(x) is negative at x = π/2, we have a local maxima which is an extrema here at x = π/2.

b. f’(2π) = cos (2π) = 1

Since we don’t have f’(x) equal to zero. We’ll not have any extrema at x = 2π.

C. f’( $\frac{3\pi }{2}$ ) = $cos\left(\frac{3\pi }{2}\right)=0$

f”( $\frac{3\pi }{2}$ ) = -sin ( $\frac{3\pi }{2}$ ) = 1

Since f”(x) > 0 we’ll have a local minima here which is also an extrema at x = $\frac{3\pi }{2}$.