When a tangent slope does not exist

Try

Sketch f(x)=|x| near zero. It has no hole or jump. Draw a line from the origin to a point on its left branch, then to a point on its right branch. Do those lines have the same slope?

Predict

Predict whether continuity guarantees differentiability. Then predict whether multiplying |x| by x changes the answer at zero. Do not judge only from whether a graph looks smooth at a particular zoom level.

Experiment

Since f(0)=0, the difference quotient for |x| at zero is |h|/h. For the comparison function g(x)=x|x|, it is h|h|/h=|h|, for nonzero h.

h	Quotient for ∣x∣	Quotient for x∣x∣
−0.1	−1	0.1
−0.01	−1	0.01
0.01	1	0.01
0.1	1	0.1

The first quotient has different one-sided limits, so no derivative exists at zero. The second approaches zero from both sides, giving derivative zero. Both functions are continuous at zero. Use the formulas to establish the behavior for every nearby input, not just these four rows.

Understand

Differentiable at a means the finite derivative f′(a) exists. A corner fails because the left and right slope limits differ. A vertical tangent can fail because the slopes are unbounded: for the real cube-root function at zero, the quotient is 1/h^(2/3), which grows positively without bound on both sides. We describe a vertical tangent, not a derivative equal to a real number called infinity.

Differentiability implies continuity. To see why, rewrite the increment as f(a+h)−f(a)=h times [f(a+h)−f(a)]/h. If the quotient has a finite limit and h approaches zero, their product approaches zero. Therefore f(a+h) approaches f(a). A jump or removable discontinuity cannot have a derivative at the discontinuous point.

The converse is false, as |x| demonstrates. Continuity controls the size of the output increment; differentiability also controls how that increment compares with the input increment.

For a piecewise join, check value matching before slope matching. Suppose f(x)=x² for x≤1 and f(x)=ax+b for x>1. The actual value is 1. Continuity requires a+b=1. The left slope at the join is 2; the right slope, after value matching, is a. Thus differentiability requires a=2 and b=−1. Matching slopes while allowing a jump would be invalid: the difference quotient across the jump would contain a nonvanishing output difference divided by a vanishing input difference.

Apply

Use three diagnostic questions: is the function continuous; do both one-sided quotients have finite limits; do those limits agree? Written answers should show the quotient rather than just labeling a graph “sharp.” The practice proof asks you to explain the one-way implication and provide a counterexample to its converse.

Continue

Derivative rules make calculations faster once their assumptions are respected. The next lesson uses power and linear-combination rules while retaining domain restrictions and checking misleading shortcuts.