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Sketch this function for $k = 1$. Is it continuous? Find any values of $k$ for which $f$ is continuous.

$$f(x)= \begin{cases} kx+3, & \text{$x≤1$} \\ (kx)^2-5, & \text{$x>1$} \end{cases}$$

I would imagine that for the left side, I would get $4 (x+3$, which $1$ is plugged into $x$) and for the right side, I would get $-4$ (plug in $1$ for $x$, squared $- 5$ to get $-4$). I would guess these limits don't match, as one is positive and the other is negative. As they don't match, we wouldn't have a rational function, so this function is discontinuous at $x=1$ and the discontinuity is a jump discontinuity, as the $4$ and $-4$ don't match.

Am I on the right path here? If not, where did I go wrong?

Much appreciated in advance! :)

UPDATE - Thank you all for the replies! Have a lovely week. :)

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    $\begingroup$ You are right. . $\endgroup$
    – Arjun
    May 31, 2020 at 14:42
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    $\begingroup$ Now try calculating the value(s) for k for which it is continuous (if any) $\endgroup$ May 31, 2020 at 14:47
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    $\begingroup$ Please don't use generic terms like "query" in the title. All posts on this site are questions – imagine how the main page would look and how inefficient it would be if everyone did that. $\endgroup$
    – joriki
    May 31, 2020 at 14:57

2 Answers 2

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You are totally right :), it seems that you have understood the concept very well, but in the future, when you are dealing with piecewise functions, instead of just substituting the values of $x$, you should calculate left hand limits and right hand limits. Though for this question there wasn't any need to do that. Let me give you a simple example:$$f(x)=\begin{cases} \lfloor x \rfloor & x \leq 0\\x & x>0\end{cases}$$ now if you substitute the value of $x=0$ you may think that it is continuous but no, you have to check the left hand limit which is -1 so the condition for continuity is left hand limit $=f(a)=$ right hand limit

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At $(-\infty,1)$ and $ (1,+\infty) $, $ f$ is continuous since it has polynomial form.

at $ x=1$,

$$\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(kx+3)=k+3$$

$$\lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}(k^2x^2-5)=k^2-5$$

thus, $ f$ is continuous at $x=1$ if and only if $$k+3=k^2-5$$ or $$k^2-k-8=0$$ you can finish.

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