Does $f(0)=0$ Imply That $f'(x)=0$?

No. Knowing that $f(0)=0$ tells you the function’s value at one input. It does not tell you how the function changes near that input, which is what a derivative measures. In particular, it does not imply that $f'(0)=0$, much less that $f'(x)=0$ for every $x$.

For example, take $f(x)=x$. Then $f(0)=0$, but $f'(x)=1$ for every real $x$. The graph passes through the origin and has a nonzero slope everywhere.

What the condition tells you

The equation $f(0)=0$ means that the point $(0,0)$ lies on the graph of $f$. It is a condition on the function’s height at the input $0$, not on its slope. Many functions with different slopes—and even functions with no derivative at zero—can pass through that point.

The derivative at zero, when it exists, is defined by the limit

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

Since $f(0)=0$, this becomes

$$f'(0)=\lim_{h\to 0}\frac{f(h)}{h}.$$

To evaluate this limit, you need to know how $f(h)$ behaves for inputs $h$ close to zero. The single value $f(0)$ does not provide that information.

Counterexamples: different slopes, same value

Consider the family of functions $f(x)=cx$, where $c$ is any real number. Every member satisfies

$$f(0)=c\cdot 0=0,$$

while its derivative is $f'(x)=c$. Choosing $c=0$ gives a zero derivative; choosing $c=1$ gives a derivative of $1$; choosing $c=-3$ gives a derivative of $-3$. Thus the value at zero is compatible with many different derivatives.

A nonlinear example makes the same point. If $f(x)=x^2$, then $f(0)=0$, but $f'(x)=2x$: the derivative is zero at zero, yet it is not zero at other inputs. If instead $f(x)=x^2+x$, then $f(0)=0$ and $f'(0)=1$. The extra term changes the slope without changing the value at zero.

These examples distinguish two possible readings of the question:

• If you mean “Does $f(0)=0$ imply $f'(0)=0$?”, the answer is no; $f(x)=x$ is a counterexample.
• If you mean “Does $f(0)=0$ imply $f'(x)=0$ for every $x$?”, the answer is also no; the same counterexample has derivative $1$ everywhere.

The function may not be differentiable

The condition does not even guarantee that $f'(0)$ exists. For example, let $f(x)=|x|$. Then $f(0)=0$, but the left-hand slope at zero is $-1$ and the right-hand slope is $1$. Because these one-sided slopes differ, the derivative at zero does not exist.

So it is useful to keep three questions separate: what is the function’s value, does its derivative exist, and what is the derivative’s value? Knowing the answer to the first question does not settle the other two.

When does a zero derivative imply something?

A stronger condition is that $f'(x)=0$ at every point of an interval. If $f$ is differentiable on an interval containing $0$, then the Mean Value Theorem shows that $f$ is constant on that interval. If it also satisfies $f(0)=0$, that constant must be zero, so $f(x)=0$ throughout the interval.

The direction of the implication matters:

$$f'(x)=0\text{ throughout an interval and }f(0)=0 \quad\Longrightarrow\quad f(x)=0\text{ throughout that interval}.$$

This does not reverse to say that $f(0)=0$ forces $f'(x)=0$. A single point on the graph cannot determine the behavior of the whole function.

Common mistakes

Confusing a zero value with a zero slope. The equation $f(0)=0$ says the graph crosses or touches the horizontal level $y=0$ at $x=0$. It does not say the graph is flat there. The line $y=x$ passes through the origin with slope $1$.

Assuming the derivative exists. A function can have a well-defined value at zero and still have a corner, cusp, or other failure of differentiability there. The absolute-value example has a corner.

Reading $f'(x)=0$ as a statement at one point. The notation with variable $x$ usually asserts that the derivative is zero for every input in the stated domain. A claim about only the derivative at zero would be written $f'(0)=0$. Neither claim follows from $f(0)=0$ alone.