*f *" is never undefined. The roots of *f *" are *-*2 and *-*1:
Figure out what the sign of *f *" is doing. **Way 1:** The roots of *f *" break up the numberline into intervals:
Take a number from each interval, plug it into *f *", and see if you find a positive or a negative value: If you find a positive value, *f *" will be positive for that whole interval. If you find a negative value, *f *" will be negative for that whole interval: **Way 2:** Way 2 is much like Way 1, except that we don't bother to plug in actual numbers. We still need to look at each interval, though:
When *x* < -2, the quantity (*x* + 2) will be negative and the quantity (*x* + 1) will also be negative. Therefore *f *"(*x*) will be the product of two negative numbers, and will be positive. When -2 < *x* < -1, the quantity (*x* + 2) will be positive. This is because if *x* is closer to zero than -2, adding 2 to *x* will bump it up into the positive numbers. However, (*x* + 1) will still be negative, since adding 1 to *x* will not be enough to bump it into the positive numbers. Therefore *f *"(*x*) will be the product of one negative and one positive number, therefore negative. When -1 < *x*, both (*x* + 2) and (*x* + 1) will be positive, so *f *"(*x*) will be positive. Thankfully, this gives us the same picture we had before: Since the sign of *f *" does change at both *x* = -2 and *x* = -1, both of these would be inflection points of *f*. |