Question

For what values of the constant c is the function continuous on (-infinity, +infinity)?

Answer 1

\[f(x)=\left\{ cx+1, if x \le 3 \right\} and \left\{ cx^2-1, if x >3 \right\} \] it is all supposed to be in one system of equations but I don't know how to format that.

Answer 2

looking at cx+1 and cx^2-1 by themselves
these functions are both continuous everywhere because they are polynomials
but to find c such that f is continuous everwhere
you have to find when the left limit will equal the right limit as x approaches 3

Answer 3

Right, so how would I go about doing that. It says for what values so there is more than 1. Im kinda lost.

Answer 4

have you found the left limit?

Answer 5

we are trying to solve this equation
\[\lim_{x \rightarrow 3^-}f(x)=\lim_{x \rightarrow 3^+}f(x)\]

Answer 6

what function is to the left of 3?

Answer 7

cx+1

Answer 8

\[\lim_{x \rightarrow 3^-}f(x)=\lim_{x \rightarrow 3^+}f(x) \\ \lim_{x \rightarrow 3^-}(cx+1)=\lim_{x \rightarrow 3^+}(cx^2-1) \\ \]

Answer 9

and the right function would have to be cx^2-1

Answer 10

now compute the limit that is on the left hand side of the equation

Answer 11

ok so it would be 3x+1? or really close to 3?

Answer 12

3c*

Answer 13

why do you replace c with 3 when x is the one that approaches 3

Answer 14

ok yes 3c+1

Answer 15

so the other side would be ?

Answer 16

9c-1

Answer 17

\[\lim_{x \rightarrow 3^-}f(x)=\lim_{x \rightarrow 3^+}f(x) \\ \lim_{x \rightarrow 3^-}(cx+1)=\lim_{x \rightarrow 3^+}(cx^2-1) \\ 3c+1=9c-1 \]

Answer 18

so we need to solve this linear equation

Answer 19

so there will be only one value of c

Answer 20

OH MY GOSH THANKS THAT MAKES SO MUCH SENSE. I DIDN"T REALLY UNDERSTAND WHAT IT WANTED ME TO DO.

Answer 21

np