(D) The function has a minimum at x=3 because f'(x)<0 for x<3 and f'(x)>0 for x>3.
Let's look at the available options and see why the justification is either correct or incorrect.
(A) The function has a maximum at x=5 because f'(5)=0.
* We've been told that the first derivative f'(x) is zero at x=5. That means that there's a possibility of f(5) being either a local maximum or a local minimum. But that's not a guarantee of either happening since it's possible for the function to simply have a slope of 0 for a brief instant while the function either increases or decreases. In any case, the fact that the first derivative is 0 is a necessary, but not sufficient condition for a local maximum to occur at that point. So this is a bad choice.
(B) The function has a maximum at x=5 because f'(x)<0 for x<5 and f'(x)>0 for x>5.
* Let's look at the values. We've been told that f'(x) < 0 for x<5. That means that the value of the function is decreasing since the slope is negative. And for f'(x) > 0 for x > 5 means that the slope then increases. That would indicate that f(5) is a local minimum, not a local maximum. So this is also a bad choice.
(C) The function has a minimum at x=3 because the tangent line at x=3 is horizontal.
* Having the slope being zero at a point means that one of 3 things is happening at that point. 1. The point is a local minimum, 2. The point is a local maximum. 3. The point is an inflection point. Since there's multiple possibilities, that means we can't know that the actual situation is one of the point being a local minimum without more information. So this too is a bad choice.
(D) The function has a minimum at x=3 because f'(x)<0 for x<3 and f'(x)>0 for x>3.
* Since the slope of f(x) is negative for x < 3, that means that the value is decreasing over that range. And then the slope becomes positive for x>3, so the value is increasing. Therefore the local minimum value happens at x=3, which agrees with the statement. So this is the correct choice.
(E) The function has a minimum at x=3 because f"(3)<0.,
* OK, we've been given the sign of the second derivative. Since it's negative, that indicates that the value of the first derivative is decreasing. But it doesn't give us any information about the first derivative having a value of 0, so we don't know if there's anything special about f(3). So this is a bad choice.
