Question

Continuity find k (pre-calculus)!

Answer 1

so as we speak, the subfunctions are discontinued, the 1st goes close to 2 and who knows what happens it goes somewheres, the 2nd one makes it to 2.

we know that since the 2nd one makes to 2, to x = 2 that is, well, f(2) = kx, well, let's make f(2) for the 2nd one be equal to the 1st one then, if both they equate each other, that's where they meet, at x = 2.

`f(x)= \begin{cases} k^2-24x,&x > 2\\\\ kx,&x\leqslant 2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ k^2-24x~~ = ~~kx\hspace{5em}\stackrel{\textit{now let's go to f(2)}}{k^2 - 24(2)~~ = ~~k(2)}\implies k^2-48=2k \\\\\\ k^2-2k-48=0\implies (k-8)(k+6)=0\implies \boxed{k= \begin{cases} 8\\ -6 \end{cases}}`