Continuous function: find "a" that makes piecewise fcn continuous

[Arne]

Don't know where to post this. I'm not accustomed to English math terms and what subjects go where.

I have a function F(x) that has this definition:

. . . . .\(\displaystyle F(x)\, =\, \begin{cases}x^2&\mbox{ for }\, x\, \leq\, 1\\2x\, +\, a&\mbox{ for }\, x\, >\, 1\end{cases}\)

I have to find the value of a (and I need to know how to find the value of a) that gives me a continous function. I don't speak the language in the text book fluently enough to understand the examples. Anyone can show me how to do and why?

 

[stapel]

I have a function F(x) that has this definition:

. . . . .\(\displaystyle F(x)\, =\, \begin{cases}x^2&\mbox{ for }\, x\, \leq\, 1\\2x\, +\, a&\mbox{ for }\, x\, >\, 1\end{cases}\)

I have to find the value of a (and I need to know how to find the value of a) that gives me a continous function.

To be "continuous" means that the two parts of the function (one part for x before 1 and the other part for x after 1) "meet up". In other words, when you do the graph, the two pieces have to match up at x = 1. This means that, when you do the graph, the two curves have no gap between their ends at x = 1. Yes, only the one rule really "counts" for x = 1, but you can always extend the line for the other rule to (and even through) x = 1 to see where it would have gone, had the function only had the one (second) rule.

So do the algebra to see what you need to get the same y-values for the x-value x = 1 for each of the two parts at their meeting point:

. . .x = 1:

. . . . .left half's y-value: 1
. . . . .right half's y-value: 2 + a

What value must a have in order to have the two halves equal at x = 1?

 

[Arne]

Aha, so if i understand correctly.
F(x) is continuous when both conditions meet up, meaning they have the same y value for the same value of x. And x = 1 I get because thats the breaking point in the definiton. So 1² = 2 + a → a = -1.

 

[stapel]

Aha, so if i understand correctly.
F(x) is continuous when both conditions meet up, meaning they have the same y value for the same value of x. And x = 1 I get because thats the breaking point in the definiton. So 1² = 2 + a → a = -1.

Exactly! Good work!

Note: Your calculus course will likely continue on to giving you piecewise functions, similar to this, for which they're wanting the original function to be differentiable at the break-point (in this case, x = 1). In that case, you'll be needing both the function halves, and also the derivative-function halves, to "meet up" (that is, to match values) at the break-point. Not only will the function, f(x), have to have the same y-values at the break-point's x-value; but the derivative values will have to match, too.

Have fun!