Describe The X Values At Which F Is Differentiable

You're staring at a function — maybe it's a messy piecewise thing from a homework problem, or a rational function with a denominator that vanishes at x = 2 — and the question hits: where is this thing actually differentiable?*

Not continuous. Differentiable. There's a difference, and it matters.

What Differentiability Actually Means

Let's start with the definition, but in plain English.

A function f is differentiable at x = a if the derivative f'(a) exists. That means the limit

lim (h→0) [f(a+h) - f(a)] / h

exists and is finite. Geometrically? The graph has a well-defined tangent line at that point. Here's the thing — no sharp corners. Day to day, no cusps. In practice, no vertical tangents. Worth adding: no jumps. Just a smooth curve that you could zoom in on until it looks like a straight line.

Here's the thing most textbooks bury: **differentiability implies continuity, but continuity does not imply differentiability.In practice, the reverse? ** If f is differentiable at a, it's automatically continuous at a. False. The absolute value function f(x) = |x| is continuous everywhere but fails to be differentiable at x = 0. That corner is real, and it breaks the derivative.

The Big Three Ways Differentiability Fails

Before we hunt for x-values where f is differentiable, it's faster to learn where it isn't*. Three classic culprits show up again and again.

Corners and Cusps

A corner is where the left-hand and right-hand derivatives exist but disagree. The classic example: f(x) = |x| at x = 0. The left derivative is -1. Think about it: the right derivative is +1. They don't match, so the derivative doesn't exist.

A cusp is nastier. The graph comes to a sharp point, and the tangent line goes vertical from both sides — but in opposite directions. The one-sided derivatives blow up to infinity with opposite signs. Think f(x) = x^(2/3) at x = 0. No derivative.

Vertical Tangents

Sometimes the tangent line exists but it's vertical. On the flip side, slope undefined. The slope is infinite. The curve passes the vertical line test beautifully, but the tangent at the origin is the y-axis. In real terms, the derivative doesn't exist as a real number. But f(x) = x^(1/3) at x = 0 is the standard example. Not differentiable.

Discontinuities

This one's obvious but worth stating: if f isn't continuous at a, it can't be differentiable there. Jump discontinuities, removable holes, infinite asymptotes — all of them kill differentiability. The derivative limit simply doesn't exist when the function itself has a gap or blows up.

How to Actually Find the Differentiable x-Values

Okay, so you have a function. Maybe it's given by a formula, maybe piecewise, maybe implicitly. Here's a practical workflow.

Step 1: Find the Domain

Start here. That's why f can't be differentiable where it doesn't exist. If f(x) = √(x-3), the domain is [3, ∞). Here's the thing — differentiability is only even a question on that interval. If f(x) = 1/(x-2), the domain excludes x = 2. Done — not differentiable there because it's not defined.

Step 2: Check Continuity on the Domain

On the interior of the domain, most "nice" functions (polynomials, exponentials, trig, logs on their domains) are continuous automatically. Rational functions with canceled factors? Here's the thing — absolute values? But piecewise functions? You need to verify.

For a piecewise function like:

f(x) = { x²        if x < 1
       { 2x - 1    if x ≥ 1

Check the boundary: lim(x→1⁻) f(x) = 1, lim(x→1⁺) f(x) = 1, f(1) = 1. Good. Continuous at x = 1. But that doesn't mean differentiable — see step 3.

Step 3: Compute the Derivative (Where It Makes Sense)

On open intervals where f is given by a single differentiable formula, just differentiate normally. Rational functions? Polynomials? Everywhere on their domains. sin(x), cos(x), eˣ? Differentiable everywhere. Differentiable everywhere on their domain (where denominator ≠ 0).

The tricky spots are boundaries of piecewise definitions and points where the formula changes character (absolute values, max/min functions, etc.).

At each suspicious point x = a, compute the left-hand and right-hand derivatives:

f'₋(a) = lim (h→0⁻) [f(a+h) - f(a)] / h
f'₊(a) = lim (h→0⁺) [f(a+h) - f(a)] / h

If both exist, are finite, and equal each other, then f'(a) exists and equals that common value. If they disagree, or one/both are infinite, or one doesn't exist — not differentiable at a.

Step 4: Check Endpoints of the Domain

This trips people up. If the domain is [a, b], can f be differentiable at a or b?

Standard calculus definition: the derivative at an endpoint uses a one-sided limit. So f is differentiable at the left endpoint a if the right-hand derivative exists. At the right endpoint b, if the left-hand derivative exists. Some textbooks say "differentiable on (a, b)" and "differentiable on [a, b]" to distinguish. Know which convention your class uses.

Piecewise Functions: The Real Test

Piecewise functions are where this skill gets tested. Let's work through a concrete example.

Want to learn more? We recommend why does temperature affect reaction rate and reaction of sodium hydroxide and acetic acid for further reading.

f(x) = { x² + 1     if x ≤ 2
       { 4x - 3     if x > 2

Domain: all real numbers. Continuous? Check x = 2: left limit = 5, right limit = 5, f(2) = 5. Yes.

Derivative on each piece:

For x < 2: f'(x) = 2x
For x > 2: f'(x) = 4

At x = 2: Left-hand derivative = lim(h→0⁻) [f(2+h)-f(2)]/h = lim(h→0⁻) [(2+h)²+1 - 5]/h = lim(h→0⁻) [4+4h+h²+1-5]/h = lim(h→0⁻) (4h+h²)/h = 4. Right-hand derivative = lim(h→0⁺) [4(2+h)-3 - 5]/h = lim(h→0⁺) (8+4h-8)/h = 4. Less friction, more output.

They match. f'(2) = 4. f is differentiable everywhere.

Now tweak it slightly:

g(x) = { x² + 1     if x ≤ 2
       { 5x - 5     if x > 2

Same continuity check at x = 2: g(2) = 5, right limit = 5. Continuous.

Left-hand derivative at 2: still 4. Right-hand derivative: lim(h→0⁺) [5(2+h)-5 - 5]/h = lim(h→0⁺) (10+5h-10)/

h = 5. Since 4 ≠ 5, the left and right derivatives disagree — g is not differentiable at x = 2, even though it’s continuous there. The "corner" at x = 2 breaks differentiability.

Step 5: Piecewise Functions with Discontinuities

If a piecewise function has a jump discontinuity (e.g., different left/right limits at a boundary), it’s automatically non-differentiable there. For example:

h(x) = { x³ if x < 0 { x + 2 if x ≥ 0

At x = 0:

Left limit = 0, right limit = 2, f(0) = 2 ⇒ discontinuous.
Differentiability is moot: no derivative exists here.

Step 6: Absolute Value and Max/Min Functions

Functions like |x| or max(x, −x²) have "corners" or "kinks" where derivatives fail. For |x|:

Left derivative at 0 = −1, right derivative = 1 ⇒ not differentiable at 0.
Max/min functions often have similar issues at points where the "dominant" formula switches.

Step 7: Rational Functions and Vertical Asymptotes

Rational functions (ratios of polynomials) are differentiable everywhere on their domain (denominator ≠ 0). Still, points where the denominator is zero are excluded from the domain entirely. For example:

k(x) = 1/x

Domain: x ≠ 0. Differentiable for all x ≠ 0, but undefined at 0.

Step 8: Composite Functions

For composite functions (e.g., f(g(x))), use the chain rule where g(x) is differentiable. If g(x) has a corner or discontinuity, the composite function inherits those issues. For example:

m(x) = |sin(x)|

Differentiable everywhere except at x = nπ (n ∈ ℤ), where sin(x) = 0 and the absolute value creates a corner.

Step 9: Special Cases — Cusps and Vertical Tangents

Some functions have infinite derivatives (vertical tangents) or undefined derivatives (cusps). For example:

Vertical tangent: f(x) = x^(1/3) at x = 0. The derivative f’(x) = (1/3)x^(-2/3) → ∞ as x → 0.
Cusp: f(x) = |x|^(1/3) at x = 0. Left derivative = −∞, right derivative = ∞ ⇒ not differentiable.

Step 10: Higher-Order Derivatives

If a function is differentiable everywhere, its higher-order derivatives may still fail at certain points. For example:

f(x) = x² sin(1/x) for x ≠ 0, f(0) = 0

f is differentiable everywhere (f’(0) = 0 via the squeeze theorem).
On the flip side, f’’(0) does not exist due to oscillatory behavior near 0.

Conclusion

Differentiability hinges on smoothness: no breaks, corners, cusps, or vertical tangents. While continuity is necessary, it’s not sufficient. Always:

Check continuity first (if discontinuous → not differentiable).
Compute left/right derivatives at boundaries/suspicious points.
For piecewise functions, verify derivatives match at boundaries.
Watch for infinite derivatives or undefined limits.

In the original example, f(x) is differentiable everywhere because it’s smooth at x = 1 (matching derivatives) and has no other discontinuities. For piecewise functions, even continuity doesn’t guarantee differentiability — the derivatives must align perfectly at boundaries.