For the function, use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.
We use the quotient rule to find the derivative:
This is only undefined when x2-3x + 2 = 0, but such an x value doesn't count as a critical point since it also makes f undefined. The derivative is zero when the numerator is zero, which occurs when
Now we have this numberline:
Since the denominator of f' is always positive, the sign of f' is the same as the sign of its numerator. The numerator
-6x2 + 12
is positive when 6x2 is smaller than 12, that is, when x2 is smaller than 2. When x is within sqrt 2 of 0 (that is, between -sqrt 2 and + sqrt 2),
-6x2 + 12 > 0.
-6x2 + 12 < 0.
On the numberline, we have this:
Since f' is negative to the left of x = -sqrt 2 and positive to the right of x = -sqrt 2, f has a minimum at x = -sqrt2. Since f' is positive to left of x = sqrt 2 and negative to the right, f has a maximum at x = sqrt 2.