find the constant c that makes g continuous (g in two parts)

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nona.m.nona
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find the constant c that makes g continuous (g in two parts)

Postby nona.m.nona » Mon Dec 15, 2008 2:01 pm

This is from my calculus book, but I think I only need algebra:

Find the constant c that makes g continuous on (-infinity, +infinity):

.        /
| x^2 - c^2 if x < 4
g(x) = <
| cx + 20 if x >= 4
\

(The dot above doesn't mean anything, but the first line wouldn't line up without a leading character.)

"Continous" just means that the two ends line up, right? So I just have to do:

42 - c2 = 4c + 20

16 - c2 = 4c + 20

0 = c2 + 4c + 4

0 = (c + 2)2

-2 = c

This would make the two ends meet up at (x, y) = (4, 12). Is that okay? I'm not forgetting any "calculus" stuff for this, am I?

Thanks in advance. :mrgreen:

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stapel_eliz
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Postby stapel_eliz » Tue Dec 16, 2008 12:32 pm

nona.m.nona wrote:This would make the two ends meet up at (x, y) = (4, 12). Is that okay? I'm not forgetting any "calculus" stuff for this, am I?

I'm not familiar with any extra, calc-specific stuff that you "should" do in addition to what you've done. And your value for "c" makes the two ends meet "in the middle", so your answer looks good to me! :D

Eliz.


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