For what value of c is the function
f(x)= {x^2-c^2 x less than or equal to 1}
{cx+(5/4) 1

Question

For what value of c is the function 
f(x)= {x^2-c^2  x less than or equal to 1}
         {cx+(5/4)  1<x}
continuous on (-infinity, infinity)

anonymous · Answer

This is a piecewise function

anonymous · Answer

First you need to know what continuous means. Continuous means there are no jumps or holes in the function. So this means that at the point x=1 the two parts of the piecewise function must be equal. Can you get it from here?

anonymous · Answer

I understand that, but still need some assistance in how to figure it out

anonymous · Answer

Ok just a sec and I'll give you some steps.

anonymous · Answer

Super!

anonymous · Answer

Lol I have to retype my reply since my window crashed just one more moment please.

anonymous · Answer

Thank you!

anonymous · Answer

So you set the two functions equal to each other at x=1:$$x ^{2}-c ^{2}=cx+5/4$$$$c ^{2}+cx-x ^{2}+5/4=0$$
Plugging in x=1:$$c ^{2}+c(1)+5/4-(1)^{2}=0$$$$c ^{2}+c+1/4=0$$ Solve the quadratic equation for the value(s) of c that make this piecewise function continuous. Hint this is a repeated root so you only get one value. Can you take it from here?

anonymous · Answer

c=-1/2

anonymous · Answer

Thank you SO much! Stellar explanation!!!

anonymous · Answer

Yeah that's right. Good job and good luck.

anonymous · Answer

I have  a question though... why do we set they functions equal to each other at x=1, if one function is x less than or equal to 1, and the other x>1?