Constant that makes g(x) continous over (-inf,inf)

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Homework Statement



Find the C (constant) that makes the function g(x) continuous over all real numbers.


Homework Equations



g(x) = x^2 - c^2 | if x < 4

= cx + 20 | if x >= 4

The Attempt at a Solution



Since the function is composed of a quadratic and a polynomial they are continuous over all real numbers. Thus, the only possible point of discontinuity lies at x = 4.

lim g(x) as x -> 4 = 4c + 20

I need to make the limit of g(x) as x -> 4 from the left equal to the above limit.

So logically I need to solve for c

c4 + 20 = 4^2 + c^2

Assuming I've thought this through correctly up until this point...I guess I just can't solve the equation. Do I need to use the quadratic formula?

Thanks Guys,

Jeff
 
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Quadratic formula sounds like a good place to start.
 
Actually, I believe that thinking rather than the quadratic formula is a good place to start! Go back and check your equations. You should not have 4c+ 20= 42+ c! You've dropped a sign.
 
Well, yeah, thinking is always good, but he *thought* he should try the quadratic formula and I think he's right in the sense that it will give him the correct answer.
 
It would be better to think the correct equation first!

My real point was that if he had the right equation, it would be trivial to factor.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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