Originally Posted by

**OnMyWayToBeAMathProffesor** I am sorry for the late response.

so when i first differentiate $\displaystyle f(x)=x+sin^2x$ i get $\displaystyle f'(x)=1+2\sin\cos$ which is the same as $\displaystyle f'(x)=1+\sin\cos+\sin\cos$ then when i differentiate again i get $\displaystyle f''(x)=2cos^2(x)-2sin^2(x)$ which equals 1 $\displaystyle x=\frac{\pi}{6}$ and when $\displaystyle \frac{5\pi}{6}$. how do i justify that without saying i looked at my gragh?

b)there is no minimum in the given interval of $\displaystyle \frac{\pi}{6}\le x \le \frac{5\pi}{6}$ because when f'(x)=0 that point is a maximum on the original graph. correct?

no ... f'(x) = 0 is not the sole indicator that an extrema exists there. you need to perform the 1st or 2nd derivative test to confirm it. in this case, you have endpoint extrema.

many students get confused by this true statement ... **if f has an extreme value at x, then f'(x) = 0 or f'(x) is undefined.** they want to turn it around and believe that the converse is always true ... it's not.

c) infection points occur when $\displaystyle f''(x)=0$ so that is when $\displaystyle x=\frac{\pi}{4}$ and $\displaystyle x=\frac{3\pi}{4}$

are these correct?

yes, but you need to justify that claim. an inflection point occurs where f''(x) = 0 **and** f''(x) changes sign.