Sutureover and Overintruppted and Continuous

Continuing from our Derivatives article, we know that not only are derivatives one of the most important concepts we can learn in Calculus but also that we can write the derivative of a function as a function itself! There is no need to take the derivative of a function at every point where we need to know it; we can find the derivative function!

But what makes a function differentiable? Well, we know that a function is considered differentiable if its derivative exists at every point in its domain. What does this mean, exactly?

• This means that a function is differentiable wherever its derivative is defined.

• In other words, as long as we can find the derivative at every point on the graph, the function is differentiable.

So, how do we determine if a function is differentiable?

We use Limits and Continuity!

What are Limits, Continuity, and Derivatives?

To start, we recall that the limit of a function is defined as:

Say we have a function, \( f(x) \), that is defined at all values in an open interval that contains \( a \), (with the possible exception of itself). Let \( L \) be a real number. If all values of the function \( f(x) \) approach the real number \( L \) as the values of \( x \) (as long as \( x \neq a \)) approach the number \( a \), then we say that:

• The limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \).
  • In other words, as \( x \) gets closer and closer to \( a \), \( f(x) \) gets closer and stays close to \( L \).
• This is symbolically expressed as:

\[ \lim_{x \to a} f(x) = L \]

And, the definition of Continuity is:

A function, \( f(x) \), is continuous at a point, \( p \), if and only if all of the following are true:

1. \( f(p) \) exists.
2. \( \lim_{x \to p} f(x) \) exists.
   • The limits from the left and right side of the function at that point are equal.
3. \( \lim_{x \to p} f(x) = f(p) \).

If a function fails any of these conditions, then \( f(x) \) is not continuous (also called discontinuous) at point \( p \).

The expression "if and only if" is a biconditional logic statement. It means that if A is true, then B is also true, and if B is true, then A is also true.

The Derivative of a function is defined as:

Say we have a function, \( f(x) \). Its derivative, denoted by \( f'(x) \), is the function whose domain contains the values of \( x \) such that the following limit exists:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]

Finally, the definition of differentiability is:

A function, \( f(x) \), is differentiable on an open interval, \( (a, b) \), if the limit,

\[ \lim_{h \to 0} \frac{f(c+h)-f(c)}{h} \]

exists for every number, \( c\), in the open interval, \( (a, b) \).

• If \( f(x) \) is differentiable, meaning \( f'(c) \) exists, then \( f(x) \) is continuous at \( c \) in the open interval of \( (a, b) \).

The Relationship between Limits, Continuity, and Derivatives

As we can see from these definitions, limits, continuity, and derivatives are intertwined. We use limits to:

• define continuity and derivatives, and

• determine whether functions are continuous and/or differentiable.

Together, these definitions tell us that differentiability is when the slope of the tangent line to the curve equals the limit of the derivative of the function at a point.

• This suggests to us that for a function to be differentiable, it must be continuous, and its derivative must be continuous as well.

Continuity and Differentiability – Theorem

We have just stumbled upon a key implication here: differentiable functions are continuous.

What does this mean?

• Differentiable means that at every point in its domain, the derivative exists for a function.

• The only way for the derivative to exist is if the function is continuous on its domain.

• Therefore, a differentiable function must also be a continuous function.

This brings us to the theorem – differentiability implies continuity.

Differentiability Implies Continuity

Let \( f(x) \) be a function with \( a \) in its domain. If \( f(x) \) is differentiable at \( a \), then it is also continuous at \( a \).

Proof of the theorem – differentiability implies continuity.

If \( f(x) \) is differentiable at \( x = a \), then \( f'(a) \) exists and the following limit exists:

\[ f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \]

Remember, this is the definition of the derivative!

To prove this theorem, we need to show that \( f(x) \) is continuous at \( x = a \) by showing that \( \lim_{x \to a} f(x) = f(a) \). So,

1. Begin with\[ \lim_{x \to a} (f(x) - f(a)) \]
2. Multiply and divide by \( (x - a) \) to get\[ \lim_{x \to a} \left( (x - a) \cdot \frac{f(x)-f(a)}{x-a} \right) \]
3. Use the product law for limits.\[ \left( \lim_{x \to a} (x - a) \right) \left( \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \right) \]
4. Evaluate the first limit and rewrite the second limit as \( f'(a) \) since it is the definition of the derivative.\[ 0 \cdot f'(a) \]
   1. Therefore,\[ 0 \cdot f'(a) = 0 \]
   2. This tells us that\[ \lim_{x \to a} (f(x) - f(a)) = 0 \]
5. Use the difference law for limits on the equation from step 4b.\[ \lim_{x \to a} f(x) - \lim_{x \to a} f(a) = 0 \]
6. Since \( a \) is a constant, simplify the second limit to get\[ \lim_{x \to a} f(x) - f(a) = 0 \]
7. Add \( f(a) \) to both sides.\[ \lim_{x \to a} f(x) = f(a) \]

Since \( f(a) \) is defined and \( \lim_{x \to a} f(x) = f(a) \), we can conclude that \( f(x) \) is continuous at \( a \).

Continuity Does Not Imply Differentiability

So, if differentiability implies continuity, can a function be differentiable but not continuous?

The short answer is no. Just because a function is continuous doesn't mean its derivative is defined everywhere in its domain.

Let's look at the graph of the absolute-value function:

\[ f(x) = |x| \]

The graph of the absolute-value function, \( f(x) = |x| \) – StudySmarter Originals

We know the absolute-value function is continuous because we can draw the graph without picking up our pencil.

However, we can also see that the slope of the graph is different on the left and right-hand sides. This suggests to us that the instantaneous rate of change (the derivative) is different at the vertex.

So, is the function differentiable here?

• Let's try to take the derivative at \( x = 0 \) and find out. We can do this by taking one-sided limits, using the definition of the derivative to determine if the slope on the left and right sides are equal.

The limit of \( f(x) \) at \( x = 0 \) from the left-hand side of the graph is:

\[ \begin{align}\lim_{h \to 0^{-}} \frac{f(x+h)-f(x)}{h} &= \lim_{h \to 0^{-}} \frac{(-(0+h))-0}{h} \\&= \lim_{h \to 0^{-}} \frac{-h}{h} \\&= \lim_{h \to 0^{-}} (-1) \\&= -1\end{align} \]

The limit of \( f(x) \) at \( x = 0 \) from the right-hand side of the graph is:

\[ \begin{align}\lim_{h \to 0^{+}} \frac{f(x+h)-f(x)}{h} &= \lim_{h \to 0^{+}} \frac{(0+h)-0}{h} \\ &= \lim_{h \to 0^{+}} \frac{h}{h} \\&= \lim_{h \to 0^{+}} (1) \\&= 1\end{align} \]

Comparing these two limits, we see that the slope of the left is \( -1 \) and the slope of the right side is \( 1 \). Because the two limits do not agree, the limit does not exist.

Therefore, the absolute-value function, \( f(x) = |x| \), even though it is continuous, is not differentiable at \( x = 0 \).

This was just one example where continuity does not imply differentiability. A summary of situations where a continuous function is not differentiable include:

• If the limit of the slopes of the tangent lines to the curve on the left and right are not the same at any point, the function is not differentiable.

  • In the case of the absolute-value function, this resulted in a sharp corner of the graph at \(x=0 \). This leads us to conclude that for a function to be differentiable at a point, it must be "smooth" at that point.

• A function is not differentiable at any point where it has a tangent line that is vertical.

• A function may fail to be differentiable in more complicated ways, such as a function whose oscillations become increasingly frequent as it approaches a value.

How to Determine if a Function is Differentiable

So, how can we determine if a function is differentiable?

The quickest way to tell if a function is differentiable is to look at its graph. If it does not have any of the conditions that cause the limit to be undefined, then it is differentiable. These conditions are:

• Sharp point

  • The graph of a function with a sharp point - StudySmarter Originals

• Vertical tangent (where the slope is undefined)

  • The graph of a function with a vertical tangent – StudySmarter Originals

• Discontinuity (jump, removable, or infinite)

  • The graph of a function with a discontinuity – StudySmarter Originals

The Differences between Continuous and Differentiable

What are the differences between continuous and differentiable functions?

• Continuity is a weaker condition than differentiability.

  • For a function \( f(x) \) to be continuous at \( x = a \), the only requirement is that \( f(x)-f(a) \) converges to \( 0 \) as \( x \to a \).

  • For a function \( f(x) \) to be differentiable at \( x = a \), \( f(x)-f(a) \) must converge after being divided by \( x-a \).

    • In other words, \( \frac{f(x)-f(a)}{x-a} \) must converge as \( x \to a \).

• If a function \( f(x) \) is differentiable at \( x = a \), then it is continuous at \( x = a \) as well.

  • However, if a function is continuous at \( x = a \), it is not necessarily differentiable at \( x = a \).

• If \( \frac{f(x)-f(a)}{x-a} \) converges, the numerator converges to zero, which implies continuity.

Differentiable and Continuous Examples

Spotting where a function is not differentiable.

At which values of \( x \) is \( f(x) \) not differentiable? Why?

The graph of a function with several types of discontinuities – StudySmarter Originals

Solution:

This function is not differentiable in several places. From left to right, these are:

1. At \( x = -8 \) there is a Jump Discontinuity.
2. At \( x = -6.5 \) there is a vertical tangent.
3. At \( x = -4 \) there is a sharp point.
4. At \( x = 0 \) there is an infinite discontinuity.
5. At \( x = 2 \) there is another sharp point.
6. At \( x = 3 \) there is a Removable Discontinuity.

Is the function below continuous and differentiable at \( x = 0 \)?

\[f(x) =\begin{cases}1 & x \lt 0 \\x & x \geq 0\end{cases}\]

Solution:

First, let's graph this function.

Is the function continuous and differentiable at \( x = 0 \)? — StudySmarter Originals

1. Is the function continuous at \( x = 0 \)?

• To determine if the function is continuous at \( x = 0 \), we can either:
  • look at the graph of the function, or
  • evaluate the limit as \( x \to 0 \) of both parts of the function.
• Let's start by evaluating the limit of both parts of the function.

\[ f(x) = \lim_{x \to 0^{-}} 1 = 1 \]

\[ f(x) = \lim_{x \to 0^{+}} x = 0 \]

We take One-Sided Limits of the parts of the functions because that is where those parts of the function are valid.

Because we get different results when evaluating the limits, we know the function is not continuous at \( \bf{x = 0} \). And, if we look at the graph of the function above, we can confirm that the function is not continuous there.

2. Is the function differentiable at \( x = 0 \)?

Since the function is not continuous at \( x = 0 \), it is also not differentiable at \( \bf{x = 0} \).

Is the function below continuous and differentiable at \( x = 4 \)?

\[f(x) =\begin{cases}x^{2} & x \lt 4 \\5x-4 & x \geq 4\end{cases}\]

Solution:

First, let's graph this function.

Is the function continuous and differentiable at \( x = 4 \)? — StudySmarter Originals

1. Is the function continuous at \( x = 4 \)?

• To determine if the function is continuous at \( x = 4 \), we can either:
  • look at the graph of the function, or
  • evaluate the limit as \( x \to 4 \) of both parts of the function.
• Let's start by evaluating the limit of both parts of the function.

\[ f(x) = \lim_{x \to 4^{-}} x^{2} = \lim_{x \to 4^{-}} (4)^{2} = 16 \]

\[ f(x) = \lim_{x \to 4^{+}} (5x-4) = \lim_{x \to 4^{+}} (5(4)-4) = 16 \]

We take One-Sided Limits of the parts of the functions because that is where those parts of the function are valid.

Because we get the same result when evaluating the limits, we know the function is continuous at \( \bf{x = 4} \). And, if we look at the graph of the function above, we can confirm that the function is continuous there.

2. Is the function differentiable at \( x = 4 \)?

To determine if the function is differentiable at \( x = 4 \), we need to use the formula for the definition of a derivative at a point:

\[ f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \]

In the same way we took the limit of both parts of the function to test continuity, we will need to take the derivative of both parts of the function to test differentiability.

For the function to be differentiable at \( x = 4 \), the derivative of both parts of the function need to not only exist, but also be equal. In other words:

\[ f'_{-} (a) \mbox{ must be equal to } f'_{+} (a) \]

To take the derivative from the left, we plug in \( x^{2} \) for \( f(x) \), \( 16 \) for \( f(a) \), and \( 4 \) for \( a \) and solve.

\[\begin{align}f'_{-}(a) = \lim_{x \to a^{-}} \frac{f(x)-f(a)}{x-a} & = \lim_{x \to 4^{-}} \frac{x^2-16}{x-4} \\& = \lim_{x \to 4^{-}} \frac{(x+4)(x-4)}{x-4} \\& = \lim_{x \to 4^{-}} (x+4) \\& = 4+4 \\f'_{-}(a) & = 8\end{align}\]

To take the derivative from the right, we plug in \( 5x-4 \) for \( f(x) \), \( 16 \) for \( f(a) \), and \( 4 \) for \( a \) and solve.

\[\begin{align}f'_{+}(a) = \lim_{x \to a^{+}} \frac{f(x)-f(a)}{x-a} & = \lim_{x \to 4^{+}} \frac{5x-4-16}{x-4} \\& = \lim_{x \to 4^{+}} \frac{5x-20}{x-4} \\& = \lim_{x \to 4^{+}} \frac{5(x-4)}{x-4} \\f'_{+}(a) & = 5\end{align}\]

Therefore, while both limits exist, the function is not differentiable at \( \bf{ x = 4 } \) because the limit from the left is not equal to the limit from the right. And, if we look carefully at the graph of the function, we can confirm this discontinuity because there is a sharp point at \( x = 4 \).

Derivatives and Continuity – Key takeaways

• The limit of a function is expressed as: \( \lim_{x \to a} f(x) = L \)
• A function is continuous at point \( p \) if and only if all of the following are true:
  1. \( f(p) \) exists.
  2. \( \lim_{x \to p} f(x) \) exists, i.e., the limits from the left and right are equal.
  3. \( \lim_{x \to p} f(x) = p \).
• The definition of the derivative is the limit:
  • \( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
• We use Limits to:

  • define continuity and derivatives, and

  • determine whether functions are continuous and/or differentiable.

• Differentiability implies continuity, but continuity does not imply differentiability.
• To tell if a function is differentiable, look at its graph. If it does not have any of the conditions that cause the limit to be undefined, then it is differentiable. These conditions are:

  • sharp points,

  • vertical tangents,

  • discontinuities (jump, removable, infinite)

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Source: https://www.studysmarter.us/explanations/math/calculus/derivatives-and-continuity/

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