continuous extensions

[chileroo]

I basically need some help getting started on this problem, I'm not sure even where to begin here.

The problem is:

Define h(2) in a way that extends h(t) = (t² + 3t - 10)/(t - 2) to be continuous at t = 2.

Any help would be greatly appreciated!

 

[mmm4444bot]

Think "piecewise function" and "plugging the hole".

 

[Deleted member 4993]

I basically need some help getting started on this problem, I'm not sure even where to begin here.

The problem is:

Define h(2) in a way that extends h(t) = (t² + 3t - 10)/(t - 2) to be continuous at t = 2.

Any help would be greatly appreciated!

Is the function continuous as given?

If yes, simply state "function is already continuous at t = 2"

If not, why not? In case it is discontinuous - what type of discontinuity does it have?

 

[Someone2841]

I basically need some help getting started on this problem, I'm not sure even where to begin here.

The problem is:

Define h(2) in a way that extends h(t) = (t2 + 3t - 10)/(t - 2) to be continuous at t = 2.

Any help would be greatly appreciated!

Define $\displaystyle h(x)$ with a limit:

$\displaystyle h(t) = \lim_{x \to t}\frac{x^{2} + 3x - 10}{x - 2}$

We then have $\displaystyle h(2) = \lim_{x \to 2}\frac{x^{2} + 3x - 10}{x - 2} = 7$

We could also do this:

\(\displaystyle h(x) =
\begin{cases}
\frac{t^{2} + 3t - 10}{t - 2}, & \text{if }t \ne 2 \\
7, & \text{if }t = 2
\end{cases}\)

This would actually be the same as writing $\displaystyle h(x) = t+5$. This is because $\displaystyle \frac{t^{2} + 3t - 10}{t - 2} = \frac{(t+5)(t-2)}{t - 2}$, which in turn equals $\displaystyle t+5\text{ when }t \ne 2$. But if we define $\displaystyle h(2) := 7$, we just have $\displaystyle h(x) = t+5\text{ for all }t$.