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Thread: Continuity and Differential Problem

  1. #1
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    Continuity and Differential Problem

    I'm very confused on how im suppose to prove that any K works for the graph to be continous in part A.

    Also , how do i even approach part B? I'm just kinda lost in how i should approach the whole problem!


    Here's the problem : https://imgur.com/a/POws4
    Last edited by lc99; Oct 4th 2017 at 08:00 AM.
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  2. #2
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    Re: Continuity and Differential Problem

    a) Start with the definition of continuity of a function at a point. This should enable you to write down an equation that you can solve for $k$.
    b) Again, start with the definition of the derivative at a point. This again leads to an equation that you can solve for $k$.

    If you work out how to get a), you will see the method for b).
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  3. #3
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    Re: Continuity and Differential Problem

    Well, it helps to know what "continuous" means! A function, f(x), is said to be "continuous at x= a" if and only if $\lim_{x\to a} f(x)= f(a)$. Taking the limit of f(x) as x approaches 0 from below, $lim_{x\to 0} 4x= 0$. Taking the limit of f(x) as x approaches 0 from above, $\lim_{x\to 0} kx+ 2kx^2= 0$. So the limits is 0, for any k. Further, f(0)= 0. This function is continuous at 0 for all k. And, each part is a polynomial so it is necessarily continuous for x positive or negative.

    This function is continuous for all values of k.

    For the second part, you need to know what "differentiable" means! The "derivative of f at x= a" is $\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}$. A function is "differentiable" if and only if that derivative exists. Again, every polynomial is differentiable for all x so f is clearly differentiable for all x other than 0. To find the derivative at x= 0, take the "limit from below" $\lim_{h\to 0}\frac{4(x+ h)- 4x}{h}= \lim_{h\to 0} \frac{4x+ 4h- 4x}{h}= \lim_{h\to 0} \frac{4h}{h}= 4$ and then take the "limit from above" $\lim_{h\to 0} \frac{k(x+h)+ 2k(x+ h)^2- kx- 2kx^2}{h}= \lim_{h\to 0}\frac{kx+ kh+ 2kx^2+ 4khx+ kh^2- kx- 2x^2}{h}= \lim_{h\to 0}\frac{kh+ 4khx+ kh^2}{h}= \lim_{h\to 0} k+ 4kx+ kh= k$.

    In order that f be differentiable at x= 0, those two limits must be the same: k= 4.
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    Re: Continuity and Differential Problem

    Quote Originally Posted by HallsofIvy View Post
    Well, it helps to know what "continuous" means! A function, f(x), is said to be "continuous at x= a" if and only if $\lim_{x\to a} f(x)= f(a)$. Taking the limit of f(x) as x approaches 0 from below, $lim_{x\to 0} 4x= 0$. Taking the limit of f(x) as x approaches 0 from above, $\lim_{x\to 0} kx+ 2kx^2= 0$. So the limits is 0, for any k. Further, f(0)= 0. This function is continuous at 0 for all k. And, each part is a polynomial so it is necessarily continuous for x positive or negative.

    This function is continuous for all values of k.

    For the second part, you need to know what "differentiable" means! The "derivative of f at x= a" is $\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}$. A function is "differentiable" if and only if that derivative exists. Again, every polynomial is differentiable for all x so f is clearly differentiable for all x other than 0. To find the derivative at x= 0, take the "limit from below" $\lim_{h\to 0}\frac{4(x+ h)- 4x}{h}= \lim_{h\to 0} \frac{4x+ 4h- 4x}{h}= \lim_{h\to 0} \frac{4h}{h}= 4$ and then take the "limit from above" $\lim_{h\to 0} \frac{k(x+h)+ 2k(x+ h)^2- kx- 2kx^2}{h}= \lim_{h\to 0}\frac{kx+ kh+ 2kx^2+ 4khx+ kh^2- kx- 2x^2}{h}= \lim_{h\to 0}\frac{kh+ 4khx+ kh^2}{h}= \lim_{h\to 0} k+ 4kx+ kh= k$.

    In order that f be differentiable at x= 0, those two limits must be the same: k= 4.
    I have one question about the second part. Why doesn't continuity mean that it is differentiable too? How the the last part equal K ? I get k +4kx as the limit after plugging in h for 0.
    Last edited by lc99; Oct 4th 2017 at 08:48 AM.
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    Re: Continuity and Differential Problem

    Quote Originally Posted by lc99 View Post
    I have one question about the second part. Why doesn't continuity mean that it is differentiable too? How the the last part equal K ? I get k +4kx as the limit after plugging in h for 0.
    Consider the function $f(x) = |x| = \begin{cases}-x & x \le 0 \\ x & x \ge 0\end{cases}$. It is easy to show that $f(x)$ is continuous. For all values less than zero, the derivative is negative 1. For all values greater than zero, the derivative is positive 1. What is the derivative at zero? It does not exist because the left-handed limit does not equal the right handed limit. In other words, continuity does not imply differentiability.
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  6. #6
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    Re: Continuity and Differential Problem

    "Differentiability" is a stronger condition that "continuity". If a function is differentiable at point then it must be continuous there. But, as SlipEternal showed, a function may be continuous at a point without being differentiable.
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    Re: Continuity and Differential Problem

    In otherword, the slope at x=0 should be the same for the function at x <=0 and x>0 to be differentiable?
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    Re: Continuity and Differential Problem

    Quote Originally Posted by lc99 View Post
    In otherword, the slope at x=0 should be the same for the function at x <=0 and x>0 to be differentiable?
    Consider this example:
    $f(x)=\begin{cases}x^3-2 &: x\ne 2 \\ 8 &: x= 2\end{cases}$
    Is it clear to you that $\displaystyle{\lim _{x \to {2^ + }}}f(x) = {\lim _{x \to {2^ - }}}f(x) = 6~?$
    But the function $f(x)$ does not have a derivative at $x=2$ Do you see why?
    So please rethink this post.
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  9. #9
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    Re: Continuity and Differential Problem

    Quote Originally Posted by lc99 View Post
    In otherword, the slope at x=0 should be the same for the function at x <=0 and x>0 to be differentiable?
    The problem with that wording is that, in order to have a slope "at x<= 0" and "x> 0" the function has to be linear. The derivative extends the idea of "slope" to non-linear functions. But your basic idea is correct. While the derivative of a function is not necessarily continuous, any derivative satisfies the "intermediate value property" (if f(a)= u and f(b)= v, then f(x) takes on every value between u and v at some point between a and b). So, yes, if a function is differentiable at x= a, then the limit, as x goes to a, of f'(x) is f'(a).
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