Need some help with continuity for the following piece-wise functions

Question

Let $$f(x)=\begin{cases}x^2−7 &x\leq c\\10x−32 &x>c \end{cases}$$ If $f(x)$ is continuous everywhere, then what is $c$ equal to?

Also, where is $$f(x)=\begin{cases}x+2 &x<0\\e^x &0\leq x\leq 1\\ 2−x &x>1\end{cases}$$ continuous?

Answer 1

Hint:

• Use the following for both parts, if a function $f$ is continous at $c$, then we have

$$\lim_{x \to c^-}f(x) = f(c) = \lim_{x \to c^+}f(x)$$

• Exponential and polynomials are continuous.

Answer 2

Since $f$ is continuous everywhere, it must be continuous particularly at $c$ as well. So $\displaystyle \lim_{x \to c^-}f(x)=\lim_{x \to c^+}\implies c^2-7=10c-32\implies (c-5)^2=0$ or $c=5$. For the second part, check for continuity at the intersection points $\Big($i.e. $\displaystyle \lim_{x \to c^-}f(x)=\lim_{x \to c^+}$$\Big)$ of the intervals since polynomial and exponential functions are continuous. So at $x=0$ we have $0+2\neq e^0=1$ and at $x=1$ we have $e^1\neq2-1$. Hence the interval of convergence of $f$ is $(-\infty,0)\cup(0,1)\cup(0,\infty)$

Answer 3

For the first question it's clear that each of those pieces of that function are continuous on their entire respective domains. Therefore, you just need to make sure it's continuous at the point where you switch between the two, at $x=c$. For a function to be continuous at $x=c$ you need $$\lim_{x \to c}f(x)=f(c).$$ Since $f(x)$ is different from the left and right of $x=c$, you need to show

$$\lim_{x \to c^{-}} f(x) = f(c)$$ $$\lim_{x \to c^{+}} f(x) = f(c).$$

The left sided limit would give you

$$\lim_{x \to c^{-}} (x^2-7) = c^2-7$$

Which would actually be true for any $c$ and is not very useful. However, the right sided limit would give you

$$\lim_{x \to c^{+}} (10x-32) = c^2-7.$$

Solving this equation for $c$ would tell you what value(s) for $c$ that would make the function continuous everywhere.

Here's a link to a site with a bit more detailed answer to a similar problem. It goes a bit more into why this works and why you want to solve it this way:

https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/

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The second question will be similar. All three of those pieces of that function are continuous on their own. As a result, you can say that $f(x)$ is continuous for all $x<0$, $0<x<1$, and $x>1$. You just need to show whether it is also continuous at $x=0$ and $x=1$. You will want to do this just like how I made sure the previous problem is continuous at $x=c$. The only difference is that you will get equations that don't have variables. If the equations are true then it's continuous at that point, if it's false then the function isn't continuous at that point.

For example, to show continuity at $x=0$, you need to show

$$\lim_{x \to 0^-}f(x)=f(0)$$ and $$\lim_{x \to 0^+}f(x)=f(0)$$

The right sided limit is not useful, but the left sided limit gives: $$\lim_{x \to 0^-}(x+2)=e^0$$ $$2=1$$

Since we know this is not true, $f(x)$ must not be continuous at $x=0$. Now do the same thing with $x=1$ to see if it's continuous there or not.