Why Does Continuity Not Imply Differentiability

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In this tutorial, we will explore what it means for a function to be differentiable in calculus. We will first look at the definition of differentiability. Then, we will work through several examples where we check the differentiability of various functions. Finally, we will see how differentiability is connected to continuity.

What Does it Mean for a Function to be Differentiable?

• A function f is differentiable at a if f^{\prime}{(a)} exists.
• Furthermore, it is differentiable on an open interval (a, b) [or (a, \infty) or (-\infty, a) or (-\infty, \infty) ], if it is differentiable at every number in the interval.

In other words, you can check if a function is differentiable or not by plotting its graph. If there are any issues like:

• a "corner" or a "kink" or
• discontinuities (see our Continuity at a Point tutorial for more information) or
• a vertical tangent

then the function is not differentiable at the specific point where the problem occurs.

Visually, the issues might look like this:

Problem Solving Strategy- Differentiability

When asked to determine the intervals of differentiability of a function, do the following:

• Plot the graph of the function f(x) .

• Look at the domain of the function f(x) and the possible values where it is undefined.

• Compute f^{\prime}{(x)} for each interval defined in the domain of the function at any undefined values of x .

Examples

In order to understand the concept of differentiability, let's take a look at several examples.

Example 1: Checking the Differentiability of a Function

Where is the function f(x)= |x- 3| differentiable?

In this case, we know that:

f(x)= |x- 3|= \begin{cases} x-3, \text{ if } x>=3 \\ -x+3, \text{ if } x< 3 \end{cases}

If we plot this function, we can see its graph looks like this:

We now look at the domain of f(x) ; the function f(x) is defined for all x \in \mathbb{R} .

We will skip what happens at x= 3 and just look at x> 3 and x< 3 for now.

If x> 3 , then using the definition of the limit we know that:

f^\prime(x)= \displaystyle\lim_{h \to 3}{\frac{f(x+h)- f(x)}{h}}= \displaystyle\lim_{h \to 3}{\frac{x+ h- 3- (x- 3)}{h}}

= \displaystyle\lim_{h \to 3}{\frac{x+ h- 3- x+ 3}{h}}= \displaystyle\lim_{h \to 3}{\frac{h}{h}}= 1

Therefore, f(x) is differentiable for any x>3 .

Now, let's consider what happens when x< 3 . In this case,

f^\prime(x)= \displaystyle\lim_{h \to 3}{\frac{f(x+h)- f(x)}{h}}= \displaystyle\lim_{h \to 3}{\frac{-(x+ h)+ 3- (-x+ 3)}{h}}

= \displaystyle\lim_{h \to 3}{\frac{-x- h+ 3+ x- 3}{h}}= \displaystyle\lim_{h \to 3}{\frac{-h}{h}}= -1

As a result, f(x) is differentiable for any x< 3 .

Now, let's investigate what happens at x= 3 .

We will calculate the right and left hand limits separately.

\displaystyle\lim_{h \to 3^+}{\frac{f(x+h)- f(x)}{h}}= \displaystyle\lim_{h \to 3^+}{\frac{x+ h- 3- (x- 3)}{h}}= 1

\displaystyle\lim_{h \to 3^-}{\frac{f(x+h)- f(x)}{h}}= \displaystyle\lim_{h \to 3^-}{\frac{-(x+ h)+ 3- (-x+ 3)}{h}}= -1

As a result,

\displaystyle\lim_{h \to 3^+}{\frac{f(x+h)- f(x)}{h}} \neq \displaystyle\lim_{h \to 3^-}{\frac{f(x+h)- f(x)}{h}}

so that

\displaystyle\lim_{h \to 3}{\frac{f(x+h)- f(x)}{h}}= f^\prime(3) does not exist.

So, f(x)= |x-3 | is differentiable at all x \in \mathbb{R} , except when x= 3 .

Example 2: Checking the Differentiability of a Function

Where is the function f(x)= \sqrt{x} differentiable?

You can see the graph of this function below:

We know that f(x)= \sqrt{x} is defined for all x \in \mathbb{R} such that x>= 0 .

Let's skip for now the value of x= 0 .

If x>0 , then

f^\prime(x)= \displaystyle\lim_{h \to 0}{\frac{f(x+h)- f(x)}{h}}= \displaystyle\lim_{h \to 0}{\frac{\sqrt{x+h}- \sqrt{x}}{h}}

= \displaystyle\lim_{h \to 0}{\frac{(x+h)- x}{h \cdot [\sqrt{(x+h)}+ \sqrt{x}]}}

= \displaystyle\lim_{h \to 0}{\frac{1}{\sqrt{(x+h)}+ \sqrt{x}}}= \frac{1}{2\sqrt{x}}

But, f^\prime(0) does not exist.

Therefore, f^\prime(x) is differentiable everywhere on x \in \mathbb{R} such that x> 0 .

Example 3: Checking the Differentiability of a Function

Where is the function f(x)= x^2+ 1 differentiable?

You can see the graph of this function below:

We know that f(x)= x^2+ 1 is a polynomial of degree 2 which is defined for all x \in \mathbb{R} .

We can also differentiate f(x) to get f^{\prime}{(x)}= 2x which exists for all x \in \mathbb{R} .

Therefore, f(x) is differentiable for all x \in \mathbb{R} .

Differentiability is Linked to Continuity

Recall the concepts of Continuity at a Point and Continuity on an Interval discussed in these previous tutorials.

In addition to this, we know that:

• If f is differentiable at a , then f is continuous at a .
• However, if f is continuous at a ,then f is not necessarily differentiable at a .

In other words:

• Differentiability implies continuity.

• But, continuity does not imply differentiability.

Taking a closer look at the examples we previously discussed in this tutorial, we can see that f(x)= |x- 3| is continuous at x= 3 because there are no gaps or breaks in its graph at this point and

\displaystyle\lim_{x \to 3}{f(x)}= \displaystyle\lim_{x \to 3}{|x- 3|}= 0 = f(3)

However, f(x)= |x- 3| is not differentiable at x= 3 because its graph contains a "corner" or a "kink" at this point and f^{\prime}{(3)} does not exist. In this case, the function f(x)= |x- 3| is continuous at x= 3 , but not differentiable at x= 3 .

In the second example, the function f(x)= \sqrt{x} is continuous at x= 0 because the graph has no gaps at this point and

\displaystyle\lim_{x \to 0}{f(x)}= \displaystyle\lim_{x \to 0}{\sqrt{0}}= 0 = f(0)

Nevertheless, f(x)= \sqrt{x} is not differentiable at x= 0 since its graph contains a vertical tangent at this point and f^\prime(0) does not exist. In this case, f(x) is continuous at x= 0 , but not differentiable at x= 0 .

In example 3, the function f(x)= x^2+ 1 shows no breaks in its graph and is continuous for every x \in \mathbb{R} since it is a polynomial. Also, f(x) is differentiable for every x \in \mathbb{R} . Here, differentiability implies continuity since f(x) is both differentiable and continuous for any x \in \mathbb{R} .

Practice Problems

Now it's your turn to practice these concepts and find where the following functions are differentiable.

1. f(x)= |2x- 4|
2. f(x)= x^2+ 4x+ 3
3. Determine where f(x)= \sqrt[4]{x} is differentiable.

Practice Solutions

1. f(x) is differentiable on (-\infty, 2) \cup  (2, \infty) , but not differentiable at x= 2 .
2. f(x) is differentiable on \mathbb{R} .
3. The answer is f(x) is differentiable on (0, \infty) , but not differentiable at x= 0 .

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Source: https://mathleverage.com/differentiability/

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