Reading Differentiability from Graphs

What a Graph Can Tell You

A derivative describes a function’s local rate of change at a point. On a graph, that rate is represented by the slope of the tangent line, when the tangent has a finite slope. Looking closely at a marked point can therefore help you decide whether the function appears differentiable there.

Some graph features provide strong evidence against differentiability: a break means the function is not continuous, a sharp turn means the local direction changes abruptly, and a vertical tangent does not have a finite slope. A smooth-looking curve with one clear, nonvertical tangent suggests differentiability, but appearance alone is not always a proof.

Definition: A function $f$ is differentiable at $x=a$ if the limit $$ \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} $$ exists as a finite real number. This limit, when it exists, is $f'(a)$.

Secant Slopes and the Tangent Slope

For a nonzero change $h$, the quotient $\frac{f(a+h)-f(a)}{h}$ is the slope of the secant line through the graph points $(a,f(a))$ and $(a+h,f(a+h))$. As $h$ approaches zero, the second point moves toward the marked point. If those secant slopes approach one finite value, that value is the tangent slope and the derivative.

\text{secant slope}=\frac{f(a+h)-f(a)}{h} \quad\longrightarrow\quad f'(a)\ \text{if the limit exists as a finite number.}

This gives a useful way to interpret a graph: ask whether nearby secants seem to settle toward a single finite slope. If they do, the graph is consistent with differentiability. If the graph is broken, pointed, or becomes vertical at the marked point, the picture signals a problem with having a finite derivative.

Continuity Is a Necessary Check

A differentiable function must be continuous at the point in question. The reason follows directly from the difference quotient. For $h\ne 0$, rewrite the change in function value as

f(a+h)-f(a) = h\left(\frac{f(a+h)-f(a)}{h}\right).

If the quotient approaches a finite number as $h\to 0$, then it stays bounded near zero. The factor $h$ approaches zero, so the product approaches zero. Thus $f(a+h)\to f(a)$, which is continuity at $a$. Consequently, a graph that is not continuous at the marked point cannot represent a function differentiable there.

Key takeaway: Continuity is necessary for differentiability, but it is not sufficient. A graph can be connected at a point and still have a sharp turn or a vertical tangent there.

A Practical Graph-Reading Method

Use the following sequence when a graph has a marked point. First check whether the function is defined there and whether the graph connects to that value. If it is not continuous, stop: the function is not differentiable at that point. If it is continuous, inspect the local tangent behavior. Does the curve appear to have one clear, nonvertical direction, or does it turn sharply or become vertical?

Check the point itself.
Confirm that the graph assigns a function value at the marked input.

Check continuity.
A gap, jump, or mismatch between the plotted point and nearby graph rules out differentiability.

Inspect the local direction.
A smooth-looking curve with one finite tangent slope is consistent with differentiability; a sharp turn or vertical tangent is not.

State the strength of your conclusion.
A graph can suggest differentiability, but an exact function rule or limit calculation may be needed to establish it.

Worked Examples

Worked Example: A Smooth Point with a Horizontal Tangent

Consider the graph of $f(x)=(x-2)^2+1$, marked at $x=2$. Determine whether the graph supports differentiability there and find the tangent slope.

First find the marked point’s function value:

$$ f(2)=(2-2)^2+1=1. $$

For a nearby input $2+h$, where $h\ne 0$,

$$ f(2+h)=(2+h-2)^2+1=h^2+1. $$

Now calculate the secant slope from $(2,1)$ to $(2+h, f(2+h))$:

\frac{f(2+h)-f(2)}{h} = \frac{(h^2+1)-1}{h} = \frac{h^2}{h} = h.

As $h\to 0$, this slope approaches $0$. Therefore the derivative exists and $f'(2)=0$. The graph has a horizontal tangent at $(2,1)$. As a check, differentiating the rule gives $f'(x)=2(x-2)$, so $f'(2)=2(0)=0$, matching the limit calculation.

Worked Example: A Continuous Graph with a Sharp Turn

Consider $g(x)=|x-1|+2$, marked at $x=1$. The graph is connected there, but its direction changes sharply. Test whether a single finite tangent slope is supported by the secants.

The marked value is

$$ g(1)=|1-1|+2=2. $$

For $h>0$, $g(1+h)=|h|+2=h+2$, so the secant slope is

\frac{g(1+h)-g(1)}{h} = \frac{(h+2)-2}{h} = 1.

For $h<0$, $|h|=-h$, so $g(1+h)=|h|+2=-h+2$. The secant slope is then

\frac{g(1+h)-g(1)}{h} = \frac{(-h+2)-2}{h} = -1.

Thus nearby secant slopes on one side are $1$, while those on the other side are $-1$. They do not approach one common value as $h\to 0$, so $g'(1)$ does not exist. The graph is continuous at the marked point, but the sharp turn prevents a single tangent slope. As a check, substituting any positive $h$ into the quotient gives $1$, and substituting any negative $h$ gives $-1$; the disagreement is exact, not a rounding effect.

Worked Example: A Break in the Graph

Suppose a graph represents $p(x)=1$ for $x<0$ and $p(x)=2$ for $x\ge 0$, with the point at $x=0$ marked. Decide whether $p$ is differentiable there.

The graph assigns $p(0)=2$, because the rule for $x\ge 0$ includes zero. For inputs just to the left of zero, the graph stays at height $1$, not height $2$. The function values therefore do not approach $p(0)$ from both sides, so $p$ is not continuous at zero. Since differentiability requires continuity, $p$ is not differentiable there.

One can also see why the difference quotient cannot settle to a finite derivative. For $h<0$, $p(h)=1$ and $p(0)=2$, giving

\frac{p(0+h)-p(0)}{h} = \frac{1-2}{h} = \frac{-1}{h}.

As negative $h$ approaches zero, $\frac{-1}{h}$ grows without bound. This confirms that there is no finite derivative. The graph’s break already ruled it out by continuity, and the quotient check agrees.

Worked Example: A Vertical Tangent

Consider $q(x)=\sqrt[3]{x}$, marked at $x=0$. The graph passes continuously through $(0,0)$, but it becomes vertical there. Check whether the derivative is a finite real number.

Since $q(0)=0$, for $h\ne 0$ the difference quotient is

\frac{q(0+h)-q(0)}{h} = \frac{\sqrt[3]{h}}{h} = \frac{1}{h^{2/3}}.

For every nonzero $h$, $h^{2/3}>0$. As $h\to 0$, $h^{2/3}\to 0$, so $\frac{1}{h^{2/3}}$ grows without bound rather than approaching a finite real number. Therefore $q'(0)$ does not exist as a finite derivative. The graph’s vertical tangent matches that conclusion. As a check, taking $h=1/1000$ gives $h^{2/3}=1/100$, so the quotient is $100$; taking still smaller positive $h$ makes it larger, consistent with unbounded slopes.

What Different Graph Features Suggest

Local graph feature	What it tells you
A gap, jump, or mismatch at the marked point	The function is not continuous there, so it is not differentiable there.
A connected graph with a sharp turn	Nearby secant slopes may fail to settle on one tangent slope.
A vertical tangent	The slopes become unbounded rather than approaching a finite derivative.
A smooth-looking, nonvertical tangent	The graph is consistent with a finite derivative, but a picture alone may not prove the limit exists.

A Common Mistake: Calling a Vertical Slope a Derivative

A common error is to say that a vertical tangent has derivative “infinity.” For $q(x)=\sqrt[3]{x}$ at zero, the calculation gives $\frac{1}{h^{2/3}}$, which grows without bound. But infinity is not a finite real number, and the derivative is defined as a finite real limit. The correct conclusion is that the derivative does not exist as a finite number; the graph has a vertical tangent.

Another tempting mistake is to conclude that every connected graph is differentiable. The sharp-turn example is connected at its marked point, yet its nearby secant slopes approach different values. Continuity is a necessary check, not a guarantee of differentiability.

Why Graph Reading Matters

Graph interpretation helps you connect the geometric meaning of the derivative to its limit definition. It lets you identify where a derivative may fail before carrying out algebra, and it helps you check whether a calculated derivative makes sense. For example, a calculation claiming a finite derivative at a visible break should prompt you to revisit the function rule or the point being evaluated.

Still, a plotted curve has limited resolution. A graph can hide a small break, a narrow turn, or rapid oscillation. Use the picture to form and test a conclusion, then use the function’s definition or the difference quotient when an exact justification is needed.

Summary: At a marked point, first check continuity. If the graph is continuous, inspect whether nearby secants suggest one finite tangent slope. A break, sharp turn, or vertical tangent indicates that the derivative does not exist as a finite real number; a smooth appearance is evidence, not always proof, of differentiability.

Check Your Understanding

For each question, explain what the graph or difference quotient tells you about differentiability at the marked point.

A graph has a filled point at $(3,2)$, while the nearby curve approaches height $4$. Can the function be differentiable at $x=3$? Explain.
For $r(x)=(x+1)^2$, find the secant slope from $x=-1$ to $x=-1+h$, and determine the derivative at $x=-1$.
A continuous graph has a sharp turn at its marked point. Why is continuity alone not enough to conclude differentiability?
For $s(x)=\sqrt[3]{x}$ at zero, what happens to the difference quotient as $h\to 0$, and what does that imply about a finite derivative?
Why should a smooth-looking graph be treated as evidence for differentiability rather than automatic proof?