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Thread: General Question About Differentiable Functions

  1. #1
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    General Question About Differentiable Functions

    I have this function where f(x) = 4 when x <= 0 and f(x) = 16x+4. This graph is continuous everywhere, but is it differentiable everywhere? The point at x=0 is very sharp which means that you can't possibly find the tangent slope at this point.

    Also, if we continue to find the third derivative at these functions, we get f(x) =0 when x<=0 and f(x) = 0. What does this mean now for the function f'''(x). It's just constant, and it doesn't depend on x? What does the fourth and fifth derivative mean? Is it still differentiable?
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    Re: General Question About Differentiable Functions

    Quote Originally Posted by lc99 View Post
    I have this function where f(x) = 4 when x <= 0 and f(x) = 16x+4. This graph is continuous everywhere, but is it differentiable everywhere? The point at x=0 is very sharp which means that you can't possibly find the tangent slope at this point.
    $f(x)=\begin{cases}4 &: x\le 0 \\16x+4 &: x>0\end{cases}$

    The derivative $f'(0-)=0~\&~f'(0+)=16$ so what does tell you?
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    Re: General Question About Differentiable Functions

    It is still not differentiable because there is a gap between 0 and 16, but the derivative of f(x) = 0 when x<0 and f(x) =16 at x>=0. It's not differentiable though right? Just because you can find the derivative, it doesn't mean that the point at 0 is differentiable?
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    Re: General Question About Differentiable Functions

    Quote Originally Posted by lc99 View Post
    It is still not differentiable because there is a gap between 0 and 16, but the derivative of f(x) = 0 when x<0 and f(x) =16 at x>=0. It's not differentiable though right? Just because you can find the derivative, it doesn't mean that the point at 0 is differentiable?
    The derivative doesn't exist at that point.

    The derivative is a limit and thus to exist it must have the same value when approached from either direction.

    Plato noted this isn't true in this case and thus the limit doesn't exist at this point and so the derivative doesn't exist at this point.
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    Re: General Question About Differentiable Functions

    Thanks for the reply, but I can still find the derivative of the function in general right? So f''(0+) and f''(0-) is both 0. So, when we use the limit definition of deriviative, both are 0 now. This means that we can find the derivative at 0 and that it's differentiable? is that possible?
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    Re: General Question About Differentiable Functions

    Quote Originally Posted by lc99 View Post
    Thanks for the reply, but I can still find the derivative of the function in general right? So f''(0+) and f''(0-) is both 0. So, when we use the limit definition of deriviative, both are 0 now. This means that we can find the derivative at 0 and that it's differentiable? is that possible?
    let's look at a definition of the second derivative

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

    and

    $f^{\prime \prime}(0)=\displaystyle \lim_{h\to 0}\dfrac{f^\prime(h)-f^\prime(0)}{h}$

    what value will you use for $f^\prime(0)$ above? It doesn't exist.

    so if some level derivative of a function at a point ceases to exist, no subsequent derivatives exist at that point either.
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    Re: General Question About Differentiable Functions

    Quote Originally Posted by lc99 View Post
    Thanks for the reply, but I can still find the derivative of the function in general right? So f''(0+) and f''(0-) is both 0.
    The two parts of the function, y= 4 and y= 16x+ 4, are both straight lines so the second derivatives are 0 for x< 0 and x> 0. But that does NOT mean the second derivative exists at x= 0! The first derivative is 0 for x< 0 and 16 for x> 0 but does not exist at x= 0 so the second derivative cannot exist at x= 0.

    The first derivative of this function is "4 for x< 0, 16 for x> 0, does not exist at x= 0". The second derivative (and all higher derivatives) is "0 for x< 0 or x> 0, does not exist at x= 0".

    So, when we use the limit definition of deriviative, both are 0 now. This means that we can find the derivative at 0 and that it's differentiable? is that possible?
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    Re: General Question About Differentiable Functions

    Quote Originally Posted by HallsofIvy View Post
    The two parts of the function, y= 4 and y= 16x+ 4, are both straight lines so the second derivatives are 0 for x< 0 and x> 0. But that does NOT mean the second derivative exists at x= 0! The first derivative is 0 for x< 0 and 16 for x> 0 but does not exist at x= 0 so the second derivative cannot exist at x= 0.

    The first derivative of this function is "4 for x< 0, 16 for x> 0, does not exist at x= 0". The second derivative (and all higher derivatives) is "0 for x< 0 or x> 0, does not exist at x= 0".

    But, I defined f(x) = 16 at x>=0 including 0? Wouldn't that mean there would be a point?
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    Re: General Question About Differentiable Functions

    You originally wrote "f(x) = 4 when x <= 0 and f(x) = 16x+4". I think you meant "f(x)= 16x+ 4 for x> 0". The graph passes through the point (0, 4) and is "continuous" at x= 0 (and so for all x). But the function is not "differentiable" there. You can see that by looking at the graph of f(x) and noting that the there is a sharp angle there- the graph is not "smooth". Or you can calculate that the derivative of f, for x< 0, is 0 and for x> 0 is 16. As I said before, while the derivative of a differentiable function is NOT necessarily continuous, it does satisfy the "intermediate value property". That is, since f'(x) is 0 at, say, x= -0.0001, and f'(x)= 16 for x, say, x= 0.0001, f'(x) would have to take on all values between 4 and 16 but it doesn't. Therefore f cannot be differentiable at x= 0.
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