\begin{align}
|f(x)| &amp;= |f(x) - f(a)| = |f'(c)||x-a| \le A|f(c)||x-a| \\
&amp; = A|f(c) - f(a)||x-a| = A|f'(d)||d-a||x-a| \le A|f'(d)||x-a|^2 \\
&amp; \le |f(d)||A(x-a)|^2.
\end{align}
Now suppose there exists $c \in ]a,x[$ such that $|f(x)| \le |f(d)| |A(x-a)|^n$, by induction on $n$. Then
\begin{align}
|f(x)| &amp; \le |f(c)||A(x-a)|^n = |f(c) - f(a)||A(x-a)|^n \\
&amp; = |f'(d)||d-a||A(x-a)|^n \le A|f(d)||d-a||A(x-a)|^n \\
&amp; \le |f(d)||A(x-a)|^{n+1}. 
\end{align}
Hence for every integer $n$ there exists $c_n$ such that $|f(x)| \le |f(c_n)||A(x-a)|^{n+1}$. Since $[a,b]$ is compact and $f$ is differentiable, $f$ is continuous, hence $|f|$ is continuous and attains a maximum $M$ over $[a,b]$, so that $|f(x)| \le |f(c_n)| |A(x-a)|^n \le M|A(x-a)|^n$. This means $f(x) = 0$ as long as $|x-a| &lt; \frac 1A$, since the bound only depends on $n$, hence we can let it go to infinity. By continuity of $f$ you also get $f(x) = 0$ if $|x-a| \le \frac 1A$.

Since you can split the interval $[a,b]$ as $[a,a+\frac 1A]$, $[a+\frac 1A, a + \frac2A]$, $\dots$, $[a+\frac mA, b]$, you can repeat this process finitely many times and show that $f$ is zero over $[a,b]$. 

Hope that helps,

derivative bounded by a constant multiple of the function
Is potential energy and "work done" the same thing?

If they are not one and the same thing then why is potential energy always associated with "work done"?

Could you explain me with some examples?
