Question

The formula is given by (2x^3 +3x^2 +10x-9/(x-1)
when x<-1 and by the formula 2x^2-1x+a
when -1> or equal to -1
what does a need to be to make the function continuous at -1 12 years ago

Answer 1

What is your definition of something being continuous? Just, how you think about it and not necessarily the textbook definition.

Answer 2

have you covered piecewise functions yet?

Answer 3

that's what im confused about, I know what a continuous function is. This function is saying that it is discontinuous at -1 , but it think Im confused about what they are asking

Answer 4

one of the functions has a variable "a"
and depending on that variable's value, THAT FUNCTION will meet THE OTHER FUNCTION

Answer 5

still confused?

Answer 6

no I perfectly understand that, im just a little confused on how to go about solving. I dont thing im connecting the concept with the applicability of the problem

Answer 7

\(\bf f(x)=
\begin{cases}
\cfrac{2x^3 +3x^2 +10x-9}{x-1} & x< -1
\\ \quad \\
\bf 2x^2-1x+{\color{red}{ a}} & x \ge -1
\end{cases}
\\ \quad \\
when\quad x={\color{red}{ a}}\quad \cfrac{2x^3 +3x^2 +10x-9}{x-1}=\textit{some value}
\\ \quad \\
when\quad x={\color{red}{ a}}\quad 2x^2-1x+{\color{red}{ a}}=\textit{same exact value as above}
\\ \quad \\
\textit{thus we can say}
\\ \quad \\
\cfrac{2x^3 +3x^2 +10x-9}{x-1}=2x^2-1x+{\color{red}{ a}}\)

Answer 8

thank u!

Answer 9

yw