Solved Math Exams

What Is Differentiability?

A function is differentiable at a point if its derivative exists at that point. Geometrically, this means the function has a well-defined tangent line — no sharp corners, no jumps, no vertical tangents.

🔑 Key Insight: Differentiability is a stronger condition than continuity.

Differentiable → Always continuous

Continuous → NOT always differentiable

The Definition

A function $f$ is differentiable at $x = a$ if this limit exists:

$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

For the limit to exist, the left-hand and right-hand limits must be equal:

$$\lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$$

Where Functions Fail to Be Differentiable

1. Sharp Corners (Cusps)

At a corner, the left and right slopes are different.

Example: $f(x) = |x|$ at $x = 0$

From the left: slope = $-1$
From the right: slope = $+1$
These don't match → not differentiable at $x = 0$

2. Discontinuities

If a function isn't continuous at a point, it can't be differentiable there.

Example: A step function has a jump — no tangent line possible.

3. Vertical Tangents

When the tangent line is vertical, the derivative is undefined (infinite slope).

Example: $f(x) = \sqrt[3]{x}$ at $x = 0$ $$f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$$ As $x \to 0$, $f'(x) \to \infty$

Piecewise Functions: Finding $a$ and $b$

A common problem type: Given a piecewise function, find constants that make it differentiable everywhere.

Two conditions must be satisfied at the boundary point:

Continuity: The pieces must connect (same $y$-value)
Differentiability: The slopes must match (same derivative)

Example 1: Making a Piecewise Function Differentiable

Find $a$ and $b$ such that $f$ is differentiable everywhere:

$$f(x) = \begin{cases} ax + b & x \leq 1 \\ x^2 & x > 1 \end{cases}$$

Step 1: Set up the continuity condition at $x = 1$

For continuity, the left and right pieces must give the same value at $x = 1$:

Left piece at $x = 1$: $a(1) + b = a + b$

Right piece at $x = 1$: $(1)^2 = 1$

Continuity equation: $$a + b = 1$$

Step 2: Set up the differentiability condition at $x = 1$

For differentiability, the derivatives must match at $x = 1$:

Left piece: $\frac{d}{dx}[ax + b] = a$

Right piece: $\frac{d}{dx}[x^2] = 2x$, so at $x = 1$: $2(1) = 2$

Differentiability equation: $$a = 2$$

Step 3: Solve the system

From Step 2: $a = 2$

Substitute into $a + b = 1$: $$2 + b = 1$$ $$b = -1$$

$$\boxed{a = 2, \quad b = -1}$$

Verification: $f(x) = \begin{cases} 2x - 1 & x \leq 1 \\ x^2 & x > 1 \end{cases}$

At $x = 1$: Left gives $2(1) - 1 = 1$, Right gives $1^2 = 1$ ✓

Derivatives: Left gives $2$, Right gives $2(1) = 2$ ✓

Example 2: Quadratic Meets Cubic

Find $a$ and $b$ such that $f$ is differentiable everywhere:

$$f(x) = \begin{cases} x^2 + ax + b & x \leq 2 \\ x^3 - 4x & x > 2 \end{cases}$$

Step 1: Continuity at $x = 2$

Left: $(2)^2 + a(2) + b = 4 + 2a + b$

Right: $(2)^3 - 4(2) = 8 - 8 = 0$

Continuity equation: $$4 + 2a + b = 0$$

Step 2: Differentiability at $x = 2$

Left derivative: $\frac{d}{dx}[x^2 + ax + b] = 2x + a$, at $x = 2$: $2(2) + a = 4 + a$

Right derivative: $\frac{d}{dx}[x^3 - 4x] = 3x^2 - 4$, at $x = 2$: $3(4) - 4 = 8$

Differentiability equation: $$4 + a = 8$$ $$a = 4$$

Step 3: Solve for $b$

Substitute $a = 4$ into $4 + 2a + b = 0$: $$4 + 2(4) + b = 0$$ $$4 + 8 + b = 0$$ $$b = -12$$

$$\boxed{a = 4, \quad b = -12}$$

Example 3: Finding Points of Non-Differentiability

Find all points where $f(x) = |x^2 - 4|$ is not differentiable.

Step 1: Identify where the expression inside changes sign

$x^2 - 4 = 0$ when $x = \pm 2$

Step 2: Rewrite as a piecewise function

$$f(x) = \begin{cases} x^2 - 4 & x \leq -2 \\ -(x^2 - 4) = 4 - x^2 & -2 < x < 2 \\ x^2 - 4 & x \geq 2 \end{cases}$$

Step 3: Check derivatives at the boundary points

At $x = -2$:

Left derivative: $\frac{d}{dx}[x^2 - 4] = 2x$, so $2(-2) = -4$
Right derivative: $\frac{d}{dx}[4 - x^2] = -2x$, so $-2(-2) = 4$
$-4 \neq 4$ → Not differentiable at $x = -2$

At $x = 2$:

Left derivative: $\frac{d}{dx}[4 - x^2] = -2x$, so $-2(2) = -4$
Right derivative: $\frac{d}{dx}[x^2 - 4] = 2x$, so $2(2) = 4$
$-4 \neq 4$ → Not differentiable at $x = 2$

$$\boxed{f \text{ is not differentiable at } x = -2 \text{ and } x = 2}$$

Example 4: Exponential Meets Polynomial

Find $a$ and $b$ such that $f$ is differentiable at $x = 0$:

$$f(x) = \begin{cases} e^{ax} & x \leq 0 \\ x^2 + bx + 1 & x > 0 \end{cases}$$

Step 1: Continuity at $x = 0$

Left: $e^{a(0)} = e^0 = 1$

Right: $(0)^2 + b(0) + 1 = 1$

Both equal $1$ ✓ — continuity is automatic here!

Step 2: Differentiability at $x = 0$

Left derivative: $\frac{d}{dx}[e^{ax}] = ae^{ax}$, at $x = 0$: $ae^0 = a$

Right derivative: $\frac{d}{dx}[x^2 + bx + 1] = 2x + b$, at $x = 0$: $0 + b = b$

Differentiability equation: $$a = b$$

$$\boxed{a = b \text{ (any equal values work, e.g., } a = b = 3\text{)}}$$

The Checklist for Piecewise Differentiability

At each boundary point $x = c$:

✅ Check continuity: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$

✅ Check differentiability: $\lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x)$

If either fails, the function is not differentiable at that point.

Common Mistakes and Misunderstandings

❌ Mistake: Checking only continuity

Wrong: "The pieces connect, so it's differentiable."

Why it's wrong: Continuity is necessary but not sufficient. You must also check that the slopes match.

Example: $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there.

❌ Mistake: Forgetting to evaluate derivatives at the boundary

Wrong: Setting $2x + a = 3x^2 - 4$ (comparing derivative formulas directly)

Why it's wrong: You need to evaluate both derivatives at the specific boundary point.

Correct: If the boundary is $x = 2$, compare $2(2) + a$ with $3(2)^2 - 4$.

❌ Mistake: Assuming differentiable means smooth everywhere

Wrong: "If I find $a$ and $b$, the function is smooth."

Why it's wrong: Making a function differentiable at one point doesn't guarantee it's differentiable elsewhere. Check all boundary points.

Quick Reference

Differentiability requires:

Continuity at the point
Left derivative = Right derivative

Common non-differentiable points:

Corners/cusps (absolute value functions)
Discontinuities (jumps, holes)
Vertical tangents

For piecewise functions: Set up two equations (continuity + matching derivatives) and solve the system.

Differentiability for MATH 140

Exam Relevance for MATH 140

This skill appears on: