Question

for what values of c is the function f continuous on (-inf,inf) where...

Answer 1

\[f(y)= {y^2-c; y \in(-\inf,3)} ; {cy+1; y \in(3,\inf)} \]

Answer 2

Let's rewrite it in 'piecewise function.' \[f(y) = \left\{ \begin{matrix}y^2 - c, \space y \le 3 \\ cy + 1, \space y > 3 \end{matrix} \right.\]
So in order for f(y) to be continuous, y² - c and cy + 1 must be equal to each other at y = 3. Does this make sense?

Answer 3

yes a little

Answer 4

What part don't you understand? I'd be glad to clarify something.

Answer 5

@d92292

Answer 6

no like i understand it to the point im confuesued on the next step

Answer 7

do we set it equal to each other?

Answer 8

Yes! Then find c. Remember, we know that it is equal to each other at y = 3.

Answer 9

how do we set it up to equal each other at y=3

Answer 10

Just set it up, like that. y² - c = cy + 1 We know this is only true when y =3 so plug 3 into y.

y² - c = cy + 1
(3)² - c = c(3) + 1

Does this make sense?

Answer 11

yes

Answer 12

so we get 9-c=3c+1

Answer 13

Yeah.

Answer 14

8=4c

Answer 15

c = 2

Answer 16

thanks!
i understand it now

Answer 17

Glad I helped.