Calculus I · Unit 2A · lesson

Differentiability and Continuity

Concept

Learning objectives

Explain why differentiability implies continuity and identify common ways differentiability fails.

When a Derivative Exists

Explanation

Before the formulas

The main idea behind Differentiability and Continuity is local change. A graph may be complicated over a large interval and still behave in a simple, nearly linear way near one input. The derivative records that local direction. It does not describe the total amount of the function, and it does not automatically describe what happens far from the point.

Use three representations whenever possible: a numerical rate from nearby values, a slope on a graph, and a symbolic limit or derivative. When all three tell the same story, the calculation is much easier to trust. When they disagree, the disagreement usually exposes a dropped sign, a misread unit, or a function that is not differentiable at the point.

All four graphs are continuous at the marked point. Only the smooth graph has one finite limiting slope; the corner, cusp, and vertical tangent fail for different reasons.
Read this graph as text

Continuity does not guarantee a derivative. All four graphs are continuous at the marked point. Only the smooth graph has one finite limiting slope; the corner, cusp, and vertical tangent fail for different reasons. The graph can remain unbroken and still fail to have an ordinary derivative. At a corner, the finite one-sided slopes disagree. At a cusp, the slopes diverge with opposite signs. At a vertical tangent, the slopes become unbounded with the same overall vertical direction. Continuity asks whether the graph joins; differentiability asks whether it has one finite local direction.

The visual uses labeled positions, solid and dashed line styles, and written descriptions so continuity does not guarantee a derivative does not depend on color.

Why it matters: The gallery separates four behaviors that are often collapsed into the vague phrase "not smooth." The student should be able to diagnose the exact failure by examining one-sided secant slopes. The first panel provides a control case so the other three are not interpreted as the only possible continuous shapes.

Visual study

All four graphs are continuous at the marked point. Only the smooth graph has one finite limiting slope; the corner, cusp, and vertical tangent fail for different reasons.

A derivative exists only when nearby secant slopes approach one finite value. A jump breaks the underlying function values; a corner gives conflicting one-sided slopes; a cusp sends the slopes toward opposite infinities; and a vertical tangent may produce an infinite slope rather than an ordinary derivative.

Differentiability is stricter than continuity. A differentiable function must be continuous, but a continuous graph may still turn too sharply to have a derivative. The absolute-value graph is the standard reminder: it has no hole or jump at the origin, yet its left and right slopes disagree.

Theorem

Differentiability implies continuity

If ff is differentiable at x=ax=a, then ff is continuous at aa.

Proof idea

Write

f(x)f(a)=f(x)f(a)xa(xa).f(x)-f(a)=\frac{f(x)-f(a)}{x-a}(x-a).

As xax\to a, the quotient approaches f(a)f'(a) and xa0x-a\to0. Their product approaches 00, so f(x)f(a)f(x)\to f(a). A tangent slope cannot exist at a point where the graph is torn apart.

The converse is false. A continuous function may fail to be differentiable.

Common failure modes include:

• Discontinuity. A jump, hole, or vertical asymptote prevents differentiability. • Corner. Left and right derivatives are finite but unequal, as with x|x| at 00. • Cusp. One-sided slopes become oppositely infinite, as with x2/3x^{2/3} at 00. • Vertical tangent. Slopes become infinite with the same sign, as with x1/3x^{1/3} at 00. • Oscillation. Tangent behavior does not settle.

Guided walkthrough

Why x|x| is not differentiable at zero

Let f(x)=xf(x)=|x|. Compute the one-sided derivative quotients at 00.

Answer reveal

Worked solution

Write a real attempt before opening the supplied answer.

Interactive checkdifferentiability-01

Which statement best describes f(x)=x3f(x)=|x-3| at x=3x=3? • discontinuous; • continuous and differentiable; • continuous but not differentiable; • differentiable but not continuous.

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Show hint

Compare the left-hand and right-hand slopes at the corner.

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Optional advanced note

Why differentiability forces continuity

If f(a)f'(a) exists, then

f(x)f(a)=f(x)f(a)xa(xa).f(x)-f(a)=\frac{f(x)-f(a)}{x-a}(x-a).

As xax\to a, the quotient approaches f(a)f'(a) and the factor xax-a approaches 00, so their product approaches 00. Hence f(x)f(a)f(x)\to f(a). The proof is short because the derivative already contains exactly the difference needed to prove continuity.

After the explanation

Use the section idea

Reading lens

Watch a secant slope stabilize into a tangent slope and then generalize from one point to a derivative function.

Mental model

A derivative exists when shrinking two-point slopes settle to one finite local slope.

Decision

Choose whether the task asks for a value at one point, a full derivative function, or an estimate from data.

Common trap

Confusing the graph's height with its slope or assuming continuity automatically gives differentiability.

Check yourself

Can you move among a limit definition, tangent slope, graph estimate, and units without changing the meaning?

Source & rights

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