# Clarification of Textbook's Explanation of Second Derivative Test, Maxima., Minima.

My textbook says the following:

The second derivative can be used to determine whether a critical point is a local maximum, a local minimum, or a saddle point. Recall that on a critical point, $$f'(x) = 0$$. When the second derivative $$f''(x) > 0$$, the first derivative $$f'(x)$$ increases as we move to the right and decreases as we move to the left. This means $$f'(x - \epsilon) < 0$$ and $$f'(x + \epsilon) > 0$$ for small enough $$\epsilon$$. In other words, as we move right, the slope begins to point uphill to the right, and as we move left, the slope begins to point uphill to the left. Thus, when $$f'(x) = 0$$ and $$f''(x) > 0$$, we can conclude that $$x$$ is a local minimum. Similarly, when $$f'(x) = 0$$ and $$f''(x) < 0$$, we can conclude that $$x$$ is a local maximum. This is known as the second derivative test. Unfortunately, when $$f''(x) = 0$$, the test is inconclusive. In this case $$x$$ may be a saddle point or a part of a flat region.

Shouldn't it say $$f'(x - \epsilon) > 0$$ and $$f'(x + \epsilon) < 0$$ for small enough $$\epsilon$$? After all, $$f'(x - \epsilon)$$ is the first derivative value as we move the point rightward of $$x$$, and $$f'(x + \epsilon)$$ is the first derivative value as we move the point leftward of $$x$$?

I would greatly appreciate it if people could please take the time to clarify this.

No, you have it backwards. Assuming $$\epsilon >0$$, then subtracting $$\epsilon$$ from $$x$$ produces a value less than (i.e., to the left of) $$x$$, and adding $$\epsilon$$ to $$x$$ produces a value greater than (i.e., to the right of) $$x$$. In other words,
$$x-\epsilon < x$$ and $$x < x+\epsilon.$$
Since $$f''(x)>0$$, $$f'$$ increases near $$x$$. Furthermore, $$f'(x)=0$$. So, if $$\varepsilon>0$$, since we have $$x-\varepsilon, we have$$f'(x-\varepsilon)<0=f'(x)