\(x^2\) \(\lfloor x \rfloor \) \(|x|\) \(\frac {\sin (x)}{x}\)
We see that if a function is differentiable at a point, then it must be continuous at
that point.
There are connections between continuity and differentiability.
Differentiability Implies Continuity If \(f\) is a differentiable function at \(x = a\), then \(f\) is
continuous at \(x=a\).
To explain why this is true, we are going to use the following
definition of the derivative
\[ f'(a) = \lim _{x\to a} \frac {f(x)-f(a)}{x-a}. \]
Assuming that \(f'(a)\) exists, we want to show that \(f(x)\) is continuous at \(x=a\), hence we must show that
\[ \lim _{x\to a} f(x) = f(a). \]
Starting with \[ \lim _{x\to a} \left (f(x) - f(a)\right ) \]
we multiply and divide by \((x-a)\) to get \begin{align*} &= \lim _{x\to a} \left ((x-a)\frac {f(x) - f(a)}{x-a}\right ) \\ &= \left (\lim _{x\to a} (x-a) \right ) \left (\lim _{x\to a}\frac {f(x) - f(a)}{x-a}\right ) &\text {Limit Law.} \\ &= \answer [given]{0}\cdot f'(a) = \answer [given]{0}. \end{align*}
Since
\[ \lim _{x\to a}\left (f(x) - f(a)\right ) = 0 , \]
we apply the Difference Law to the left hand side \[ \lim _{x\to a}f(x) - \lim _{x\to a}f(a) = 0 , \]
and use continuity of a
constant to obtain that \[ \lim _{x\to a}f(x) - f(a) = 0 . \]
Next, we add \(f(a)\) on both sides and get that \[ \lim _{x\to a}f(x) = f(a). \]
Now we see that \(\lim _{x\to a} f(x) = f(a)\),
and so \(f\) is continuous at \(x=a\). This theorem is often written as its contrapositive:
If \(f(x)\) is not continuous at \(x=a\), then \(f(x)\) is not differentiable at \(x=a\).
Thus from the theorem above, we see that all differentiable functions on \(\mathbb R\) are continuous on \(\mathbb R\). Nevertheless there are continuous functions on \(\mathbb R\) that are not differentiable on \(\mathbb R\).
Which of the following functions are continuous but not differentiable on \(\mathbb R\)?
Consider
\[ f(x) = \begin{cases} x^2 &\text {if $x<3$,}\\ mx+b &\text {if $x\ge 3$.} \end{cases} \]
What values of \(m\) and \(b\) make \(f\) differentiable at \(x=3\)? To start, we know that \(f\)
must be continuous at \(x=3\), since it has to be differentiable there. We will start by
making \(f\) continuous at \(x=3\). Write with me: \begin{align*} \lim _{x\to 3^-} f(x) &= \answer [given]{9}\\ \lim _{x\to 3^+} f(x) &= \answer [given]{m 3 + b}\\ f(3) &= \answer [given]{m 3 + b} \end{align*}
So for the function to be continuous, we must have
\[ m\cdot 3 + b =9. \]
We also must ensure that the
function is differentiable at \(x=3\). In other words, we have to ensure that the following
limit exists
\[ \lim _{x\to 3}\frac {f(x)-f(3)}{x-3}.\\ \]
In order to compute this limit, we have to compute the two one-sided limits \[ \lim _{x\to 3^{+}}\frac {f(x)-f(3)}{x-3}\\ \]
and \[ \lim _{x\to 3^{-}}\frac {f(x)-f(3)}{x-3},\\ \]
since \(f(x)\) changes expression at \(x=3\). Write with me \begin{align*} \lim _{x\to 3^{-}}\frac {f(x)-f(3)}{x-3}&= \lim _{x\to 3}\frac {x^2 -9}{x-3}\\ &= \lim _{x\to 3^{-}}\frac {\cancel {(x-3)}(x+3)}{\cancel {x-3}}\\ &= \lim _{x\to 3^{-}}\left (x+3\right )\\ &=6, \end{align*}
and \begin{align*} \lim _{x\to 3^{+}}\frac {f(x)-f(3)}{x-3}&= \lim _{x\to 3}\frac {mx+b -(3m+b)}{x-3}\\ &= \lim _{x\to 3^{+}}\frac {mx-3m}{x-3}\\ &= \lim _{x\to 3^{+}}\frac {m\cancel {(x-3)}}{\cancel {x-3}}\\ &= \lim _{x\to 3^{+}}m\\ &=m. \end{align*}
Hence, we must have
\[ m=6. \]
Ah! So now the equation that must be satisfied \begin{align*} 9 &= m\cdot 3 + b,\\ becomes\hspace {0.3in} 9 &= 6\cdot 3 + b.\\ \end{align*}
Therefore, \(b=\answer [given]{-9}\). Thus setting \(m=\answer [given]{6}\) and \(b=\answer [given]{-9}\) will give us a function that is differentiable (and hence continuous) at \(x=3\).
Can we tell from its graph whether the function is differentiable or not at a point \(a\)?
What does the graph of a function \(f\) possibly look like when \(f\) is not differentiable at \(a\)?
The function in figure A is not continuous at \(a\), and, therefore, it is not differentiable
there.
2025-01-06 19:44:21 Each of the figures A-D depicts a function that is not differentiable at \(a=1\).
![[Picture]](digInDifferentiabilityImpliesContinuity-65c56759e745a683a4315f89471a4bfc.svg) | ![[Picture]](digInDifferentiabilityImpliesContinuity-a1a026586dca2df6ca6ffbdccd3adf4e.svg) |
![[Picture]](digInDifferentiabilityImpliesContinuity-5ccef87413f79779da516e49afbad756.svg) | ![[Picture]](digInDifferentiabilityImpliesContinuity-3a630aa9c8f73fe14685acb53f94cca9.svg) |
In figures \(B\)–\(D\) the functions are continuous at \(a\), but in each case the limit
\[ \lim _{x\to a} \frac {f(x)-f(a)}{x-a} \]
does not
exist, for a different reason.
In figure \(B\)
\[ \lim _{x\to a^{+}} \frac {f(x)-f(a)}{x-a}\ne \lim _{x\to a^{-}} \frac {f(x)-f(a)}{x-a}. \]
In figure \(C\)
\[ \lim _{x\to a} \frac {f(x)-f(a)}{x-a}=\infty . \]
In figure \(D\) the two one-sided limits don’t exist and neither one of them is
infinity.
So, if at the point \(a\) a function either has a ”jump” in the graph, or a corner, or what looks like a “vertical tangent line”, or if it rapidly oscillates near \(a\), then the function is not differentiable at \(a\).