[(f(x+h) - f(x))/h] versus [(f(x) - f(0))/x]

  • Thread starter LearninDaMath
  • Start date
In summary: So \sqrt{x}- x is differentiable everywhere except x= 0.In summary, to determine if a function is differentiable on a given interval, one can use the generic formula for the derivative and look for places where it fails. In the case of f(x) = (√x) - 2x on the interval [0,1], the standard formula for its derivative fails at x=0 due to the presence of a square root term. However, by rationalizing the numerator and taking the limit, it can be shown that f(x) is differentiable everywhere on the interval except at x=0. Alternatively, using known derivative laws, it can be argued that f(x) is differentiable everywhere on the
  • #1
LearninDaMath
295
0
I have this problem, f(x) = (√x) - 2x On some interval [0,1].

One of the things I am supposed to do is to show that it's differentiable on the interval.

Now, I am aware that the definition of the derivative is f'(x) = lim as h[itex]\rightarrow[/itex] 0 of [itex]\frac{f(x+h) - f(x)}{h}[/itex] and that this formula is used to produce the derivative function of f(x).

And that replacing the x with a known x value into the formal definition should produce a derivative value at given x value.. so that using this definition of the derivative would tell me whether or not a function at a single x value is differentiable.

First Question, is what I said thus far correct?

Given what I said above is correct,

I don't want to find the derivative of f(x) = (√x) - 2x at anyone single value. I want to find whether or not it is differentiable on the interval [0,1].

How does the formula lim as x→[itex]x_{0}[/itex] of f(x) - f([itex]x_{0}[/itex])/ x - [itex]x_{0}[/itex] come into play and how does it relate to the above definition of a derivative formula?
 
Physics news on Phys.org
  • #2
Let [itex]x-x_0 = h[/itex]. Your second formula then becomes your first.
 
  • #3
So they are the same thing? And should produce the same results?
 
  • #4
LearninDaMath said:
I have this problem, f(x) = (√x) - 2x On some interval [0,1].

One of the things I am supposed to do is to show that it's differentiable on the interval.

Now, I am aware that the definition of the derivative is f'(x) = lim as h[itex]\rightarrow[/itex] 0 of [itex]\frac{f(x+h) - f(x)}{h}[/itex] and that this formula is used to produce the derivative function of f(x).

And that replacing the x with a known x value into the formal definition should produce a derivative value at given x value.. so that using this definition of the derivative would tell me whether or not a function at a single x value is differentiable.

First Question, is what I said thus far correct?

This is correct. Replacing the general x with x = a, gives us:

[tex]lim_{h \to 0} \frac{f(a+h)-f(a)}{h}[/tex]

If this limit exists, the derivative is defined for x = a.

LearninDaMath said:
How does the formula lim as x→[itex]x_{0}[/itex] of f(x) - f([itex]x_{0}[/itex])/ x - [itex]x_{0}[/itex] come into play and how does it relate to the above definition of a derivative formula?

[itex]\lim_{x \to a} \frac{f(x)-f(a)}{x-a}[/itex] is equivalent to [itex]lim_{h \to 0} \frac{f(a+h)-f(a)}{h}[/itex]. In the first case, you're basically taking the limit of the slope of lines that get closer and closer to a, but both limits are equivalent. If the limit exists, the derivative is defined at x = a.
 
  • #5
So to find where f(x)= √x - 2x is differentiable , given the interval [0,1], what do I do?

I could say 2x is differentiable everywhere because I know its graph is a line. And √x exists at x>0 but how can I confirm its differentiability?

Do I set up f'x = lim h→0 of ((√(x+h) - 2(x+h)) - (√(x) - 2x))/h and evaluate the formula to come up with the general derivative function for the original function?

Or f'x = lim x→a of ((√(x) - 2x) - (√(a) - 2a))/ x - a and altervatively evaluate this version to come up with a general derivative function for the original function?

or

should I make "x" value from the first formula equal some value from the interval...or let "a" equal some specific value?

Am I supposed to know where its differentiable already, and then simply choose that undifferentiable value into the formula to prove its undifferentiability?
 
Last edited:
  • #6
In order to determine whether or not a function is differentiable on a set, look at the generic formula for the derivative and look for places where that fails.

Here, f(x)= x1/2- x. The standard formula for its derivative at [itex]x= x_0[/itex] is
[tex]\lim_{h\to 0}\frac{(x_0+ h)^{1/2}- (x_0+h)- x_0^{1/2}+ x_0}{h}= \lim_{h\to 0}\frac{(x_0+ h)^{1/2}- x_0^{1/2}- (x_0+ h- x_0)}{h}[/tex]
[tex]= \lim_{h\to 0}\frac{(x_0+h)^{1/2}- x_0^{1/2}- h}{h}[/tex]
It should be clear that the "-h" on the end will give -1 so we need to focus on the square root.

Rationalize the numerator by multiplying both numerator and denominator by [itex](x_0)^{1/2}+ x_0^{1/2}[/itex] to get
[tex]\frac{x_0+ h- x_0}{h((x_0+h)^{1/2}+ x_0^{1/2})}[/tex][tex]= \frac{h}{h((x_0+ h)^{1/2}+ x_0^{1/2})}[/tex][tex][/tex][tex]= \frac{1}{(x_0+ h)^{1/2}+ x_0^{1/2}}[/tex]
Now that the h is gone from the denominator, we can take the limit by setting h equal to 0. That gives
[tex]\frac{1}{2\sqrt{x_0}}[/tex]
As long as [itex]x_0\ne 0[/itex].

So this is differentiable every where except at x= 0.

If you already know the derivative laws and are not required to use the defining formula (which I did only because you mentioned it specifically) you could argue that the derivative of [itex]\sqrt{x}= x^{1/2}[/itex] is [itex](1/2)x^{1/2- 1}= (1/2)x^{-1/2}[/itex] and the derivative of x is, of course, 1, so the derivative of [itex]\sqrt{x}- x[/itex] is [itex](1/2)x^{-1/2}- 1[/itex]- which is defined for all x except x= 0.
 

1. What is the difference between [(f(x+h) - f(x))/h] and [(f(x) - f(0))/x]?

The difference between [(f(x+h) - f(x))/h] and [(f(x) - f(0))/x] is the point at which the value of h is taken. In the first expression, h is taken at a point close to x, while in the second expression, h is taken at a fixed point of 0.

2. How do these expressions relate to the concept of a limit?

These expressions are related to the concept of a limit because they both represent the slope of a line between two points on a function. As h approaches 0, the value of the expressions approaches the slope of the tangent line at x, which is the definition of a limit.

3. Is there a specific scenario in which one expression would be more useful than the other?

Yes, there are scenarios where one expression would be more useful than the other. If you are trying to find the instantaneous rate of change at a specific point on a function, [(f(x+h) - f(x))/h] would be more useful. However, if you are trying to find the average rate of change over a specific interval, [(f(x) - f(0))/x] would be more appropriate.

4. Can these expressions be used for all types of functions?

Yes, these expressions can be used for all types of functions, as long as the function is continuous at the point x. If the function is not continuous, these expressions will not give an accurate representation of the slope at that point.

5. How can these expressions be applied in real-world situations?

These expressions can be applied in real-world situations to calculate the rate of change in various scenarios. For example, in physics, these expressions can be used to calculate the velocity or acceleration of an object. In economics, they can be used to calculate the marginal cost or revenue of a product. Overall, these expressions are useful in analyzing and understanding the behavior of functions in real-world situations.

Similar threads

Replies
11
Views
1K
  • Calculus
Replies
9
Views
2K
Replies
1
Views
833
Replies
2
Views
1K
Replies
3
Views
1K
Replies
1
Views
926
Replies
4
Views
844
Replies
20
Views
2K
Back
Top