f(x) = x+sin^2x?

Question

Find all values of x for which the first derivative = 1
What are the x coordinate minimums?
What are the inflection points?

Answer 1

f'(x) = 1 + 2*sin(x)cos(x) = 1 + sin(2x) as given by the previous answerer, and he or she also gave the correct values for x to have f'(x) = 1. However, he or she didn't answer your questions about minima and inflection points.

The minima occur at points where f'(x) = 0 and f''(x) is positive, and inflection points occur when f''(x) = 0. To have f'(x) = 0, you need sin(2x) = -1 ==> 2x = 3π/4 + 2kπ ==> x + 3π/8 + kπ. f''(x) = 2*sin(2x). However, we just established that these were the points at which sin(2x) = -1, which means 2*sin(2x) will also always be negative, and therefore there are no minima. This makes sense because, looking at the derivative of the function, it does not take on any negative values; the function is always increasing, and has therefore has no minima. You should be able to solve for the inflection points without much difficulty.

Answer 2

F'(x) = 1 + sin(2x)
F'(x) = 1 when sin(2x) = 0
2x = kπ
x = kπ/2