Question

Find a value of the constant k, if possible, that will make the function continuous everywhere.
https://i.gyazo.com/972948c6d441c7dde786640193bd2af7.png

Answer 1

You want to find k, such that:
\(\color{#000000}{ \displaystyle 9-(-3)^2=k/(-3) ^2 }\)

Answer 2

Because, for the function to be continuous, both parts of your piece wise function have to be connected at x=-3.

We are not concerned about x>-3, and x<-3, since there, the function is for sure continuous. (There simply isn't a reason for discontinuity when x>-3 and x<-3, because then the function is given by one part of the piece-wise or the other (but not both), and the only point where the function can be disconnected/discontinuous is x=-3.)

Answer 3

Similarly, for the second problem, you should find k, such that:
\(\color{#000000}{ \displaystyle 9-(0)^2=k/(0) ^2 }\)

Answer 4

There is a technicality here though, that is complex.

Answer 5

Technically rather, since I am given that f(0)=9
(by part 1 of the piece wise function)

therefore, I don't need the \(\color{#000000}{ \displaystyle k/x ^2 }\)
to be defined at x=0. Rather, I
need to make sure, that \(\color{#000000}{ \displaystyle k/x ^2 }\)
approaches 9 from the left side,
as x approaches 0.

That is you should find k such that.
\(\color{#000000}{ \displaystyle 9=\lim_{x\to 0}~(k/x ^2) }\)

The above notation means,
"as x approaches 0, k/x\(^2\) will tend to equal 9."

Answer 6

Well, as x tends to 0, you will just get negative infinity...

So for case 2, it is discontinuous regardless of k.
(I am telling you this, because I think that they gave you a case that is not very fair for you.)

Answer 7

Hopefully, what I said is at least partially understood.
Good luck with your maths!

(and this new notation is called "limit" if you are interested)