Piecewise Function and Mean Value Theorem.

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A function $f$ is continuous at a point $t$ if $$\lim_{\underset{x \in \text{dom}(f)}{x \to t}} f(x) = f(t)$$

For your function, when $t = 0$, the restriction to the domain of $f$ means we are not looking at any $x < 0$. That is, the condition of continuity becomes $$\lim_{x \to 0+} f(x) = f(0)$$ Which is exactly the condition you examined in (2).

When $t = 1$, both sides are in the domain, so the condition of continuity is $$\lim_{x \to 1} f(x) = f(1)$$ But for this piecewise defined function, to examine if this is true, we need to note that $\lim_{x \to 1} f(x)$ exists if and only if the two one-sided limits exist and are equal. That is, if $$\lim_{x \to 1-} f(x) = \lim_{x \to 1+} f(x)$$ which is what you examined in (1). This condition (with (2) and the similar derivative limit) was enough to determine what $a, b, c$ had to be.

But that alone doesn't show that the $f$ defined by those values of $a,b,c$ is actually continuous at $1$. For that, you must still also show that $\lim_{x \to 1} f(x)$, the common value of $\lim_{x \to 1-} f(x)$ and $\lim_{x \to 1+} f(x)$, is in fact $f(1)$ (which follows by the continuity of $x \mapsto ax+b$). I believe the intent of the problem was only to find $a,b,c$, so you don't need to actually make this argument. But all I know is what you wrote here.

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beingmathematician
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beingmathematician

Updated on January 25, 2021

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  • beingmathematician
    beingmathematician almost 3 years

    Here's the function: \begin{align} f(x) = \left\{ \begin{array}{cc} 3 & \hspace{5mm} x=0 \\ ax+b & \hspace{5mm} 0<x\leq 1 \\ x^2+3x+c & \hspace{5mm} 1<x\leq 4 \\ \end{array} \right. \end{align} Find the constraints a,b and c so that $f(x)$ satisfies the conditions of MVT. Here's how I solved. First of all I write the contuinity condition of MVT for $x=1$ and $x=0$ \begin{align} \lim_{x\to\ 1^+} f(x)=4+c \end{align} \begin{align} \lim_{x\to\ 1^-} f(x)=a+b \end{align} \begin{align} a+b=4+c \hspace{5mm}(1) \end{align} And also we can get second limit which only goes one way. \begin{align} \lim_{x\to\ 0^+} f(x)=b=f(0) \hspace{5mm} (2) \end{align} \begin{align} b=3 \end{align} And then I tried to use derivatives for finding other variables. By then, I write teh expressions and I get those: \begin{align} \lim_{\Delta x\to\ 0^{+}} \dfrac{f(1+ \Delta x)-f(1)}{\Delta x}=a \end{align} \begin{align} \lim_{\Delta x\to\ 0^{-}} \dfrac{f(1+ \Delta x)-f(1)}{\Delta x}=5 \end{align} Clearly, $a=5$ and from $(1)$ if we plug all the values that we found we get $c=4$. Here is my questions: 1-)Is it true? 2-)Are we allowed to do the evaluation that I've done in 2. More explicitly, can I equate just one side to the actual value? Because in this question we don't have any "smaller than 0" part. Thanks for any help.

    • user2661923
      user2661923 almost 3 years
      Suggested edit: try reading your query cold as if you don't know anything about what the OP (yourself) is asking. How far do you have to get into the query before you (as an outside reader) understand what the OP is asking? If you think that your query lacks immediate clarity, you are welcome to edit your own query.
    • beingmathematician
      beingmathematician almost 3 years
      I thought that I can shortened the solution.
    • user2661923
      user2661923 almost 3 years
      After presenting a function, you state: "Here's how I solved." At this point, the reader is asking himself: solved what? What is the problem that needs to be solved and what are the constraints on the problem? The reader is left to reverse-engineer the answer to those two questions by reading further. It really is a good idea to proof-read your query.
    • beingmathematician
      beingmathematician almost 3 years
      Okay I understand. Thank you!