It is not x that is undefined at 0; it is ƒ' that is undefined at 0.
Or one can say ƒ'(x) is undefined at x = 0.
If ƒ(x) = (3/2) x2/3 + x then ƒ'(x) = x–1/3 = (1/x1/3 + 1), i.e. the first term is 1 over the cube root of x. When x = 0 then the cube root of x is 0, so the fraction is undefined. There is also one point at which the derivative is 0, and that is when x = –1.
In order to ascertain on which intervals ƒ' is positive and on which it is negative, one can begin by observing that in the expression
(1/x1/3)+ 1,
the common denominator is x1/3, so the expression becomes this single fraction:
(1 + x1/3) / x1/3.
Then observe that the denominator is positive or negative according as x is positive or negative, and the numerator is positive if x > –1 and is negative if x < –1.
Therefore we consider three intervals:
(–∞, –1], [–1,0], [0,+∞),
i.e. numbers less than or equal to –1, numbers between –1 and 0 (inclusive) and numbers greater than or equal to 0.
On the first interval the numerator and the denominator are both negative, so ƒ' (not ƒ) is positive.
On the second interval the numerator is positive and the denominator is negative, so ƒ' (not ƒ) is negative.
On the third interval, the numerator and the denominator are both positive, so ƒ' (not ƒ) is positive.
Therefore ƒ (not ƒ') is increasing on the first interval, decreasing on the second and increasing on the third.
Since ƒ is continuous at the boundary points between those intervals, we conclude that ƒ has a local maximum at –1 and a local minimum at 0.
That the local maximum is not a global maximum can be seen by observing that
ƒ(x) = (3/2) x2/3 + x > x → +∞ as x → +∞.
It is only slightly more work to show that ƒ(x) → –∞ as x → –∞, so the local minimum cannot be a global minimum.
If you're drawing the graph, notice that the tangent lines approach the vertical as x approaches 0, so there is a very sharp point at x = 0.
