Need help, are these functions differentiable?

In summary, the conversation discusses whether certain functions are differentiable at certain points and explores various concepts related to derivatives, such as the definition of the derivative and the intermediate value property. One function is found to be not differentiable at a point due to discontinuity, while another is not differentiable due to the limit not existing. The conversation also includes a helpful tip about evaluating limits.
  • #1
FallArk
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I want to figure out whether the functions are differentiable at c. I think I should use some of the trig identities, but I'm not sure which ones. Any tips?
 

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  • #2
FallArk said:
I want to figure out whether the functions are differentiable at c. I think I should use some of the trig identities, but I'm not sure which ones. Any tips?

Hey FallArk! (Smile)

Let's start with the proper definition of the derivative.
That is:
$$f(x)=\begin{cases}\tan x &\text{if } -\frac\pi 2 \le x <\frac \pi 3 \\
x^2 & \text{if } x \ge \frac \pi 3
\end{cases} \quad\Rightarrow\quad
f'(\frac\pi 3) = \lim_{h\to 0} \frac{f(\frac\pi 3+h)-f(\frac\pi 3)}{h}
$$
Now let's look at the lower limit:
$$\lim_{h\to 0^-} \frac{f(\frac\pi 3+h)-f(\frac\pi 3)}{h}
= \lim_{h\to 0^-} \frac{\tan(\frac\pi 3+h)-(\frac\pi 3)^2}{h}
\to \frac{\sqrt 3 - \frac{\pi^2}{9}}{0^-} = -\infty
$$
It's not convergent is it? (Wondering)
 
  • #3
Hi FallArk,

The function in part (c) is not differentiable at $x = c$, for it is discontinuous at that point. Just check the one-sided limits.

The function in part (f), although continuous at $x = c$, is not differentiable at $x = c$; the limit $\lim\limits_{x\to 0} \dfrac{f(x)}{x}$ does not exist. Indeed, consider the sequences $x_n = \dfrac{2}{(4n+1)\pi}$ and $y_n = \dfrac{2}{(4n-1)\pi}$. Then $x_n, y_n \to 0$, but $$\lim\limits_{n\to \infty} \frac{f(x_n)}{x_n} = 1 \neq -1 = \lim\limits_{n\to \infty} \frac{f(y_n)}{y_n}$$
 
  • #4
I like Serena said:
Hey FallArk! (Smile)

Let's start with the proper definition of the derivative.
That is:
$$f(x)=\begin{cases}\tan x &\text{if } -\frac\pi 2 \le x <\frac \pi 3 \\
x^2 & \text{if } x \ge \frac \pi 3
\end{cases} \quad\Rightarrow\quad
f'(\frac\pi 3) = \lim_{h\to 0} \frac{f(\frac\pi 3+h)-f(\frac\pi 3)}{h}
$$
Now let's look at the lower limit:
$$\lim_{h\to 0^-} \frac{f(\frac\pi 3+h)-f(\frac\pi 3)}{h}
= \lim_{h\to 0^-} \frac{\tan(\frac\pi 3+h)-(\frac\pi 3)^2}{h}
\to \frac{\sqrt 3 - \frac{\pi^2}{9}}{0^-} = -\infty
$$
It's not convergent is it? (Wondering)
I was so concentrated on getting rid of the h, I did not even see that I can just evalute it. Thanks!

- - - Updated - - -

Euge said:
Hi FallArk,

The function in part (c) is not differentiable at $x = c$, for it is discontinuous at that point. Just check the one-sided limits.

The function in part (f), although continuous at $x = c$, is not differentiable at $x = c$; the limit $\lim\limits_{x\to 0} \dfrac{f(x)}{x}$ does not exist. Indeed, consider the sequences $x_n = \dfrac{2}{(4n+1)\pi}$ and $y_n = \dfrac{2}{(4n-1)\pi}$. Then $x_n, y_n \to 0$, but $$\lim\limits_{n\to \infty} \frac{f(x_n)}{x_n} = 1 \neq -1 = \lim\limits_{n\to \infty} \frac{f(y_n)}{y_n}$$

Thanks! I get it now
 
  • #5
While the derivative, f', of a differentiable function, f, is not necessarily continuous, it does satisfy the "intermediate value property" (f'(x), for x between a and b, takes on all values between f'(a) and f'(b)). In particular, the two limits, [tex]\lim_{x\to a^-} f'(x)[/tex] and [tex]\lim_{x\to a^+} f'(x)[/tex] must be equal.
 

FAQ: Need help, are these functions differentiable?

What does it mean for a function to be differentiable?

Differentiability is a mathematical concept that describes the smoothness of a function. A differentiable function is one that has a well-defined tangent line at every point in its domain. This means that the function is continuous and has a defined slope at each point.

How can I determine if a function is differentiable?

To determine if a function is differentiable, you can use the definition of differentiability, which states that a function is differentiable if the limit of the difference quotient exists. This means that the function is differentiable if it is continuous and has a defined slope at each point in its domain.

What are some common examples of non-differentiable functions?

Some common examples of non-differentiable functions include absolute value functions, step functions, and piecewise functions with sharp corners or vertical asymptotes. These types of functions do not have a defined slope at certain points in their domain, making them non-differentiable.

What happens if a function is not differentiable?

If a function is not differentiable, it means that it is not smooth or continuous at certain points in its domain. This can make it difficult to analyze or use the function in calculations. However, the function may still have certain properties that can be studied, such as its continuity or limits.

Can a function be differentiable at some points but not others?

Yes, a function can be differentiable at some points but not others. This is often the case with piecewise functions that have different definitions for different intervals. For example, a function may be continuous and differentiable at all points except for where there is a sharp corner or a vertical asymptote.

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