# Find any values of $k$ for which $f$ is continuous

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?

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

• You are right. . May 31, 2020 at 14:42
• Now try calculating the value(s) for k for which it is continuous (if any) May 31, 2020 at 14:47
• 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. May 31, 2020 at 14:57

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
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.