Need help with calculus involving a piecewise defined function

How would one go about answering a question such as this?

f(x)= {x^2-c^2, x<4;
{cx+20,x>=4

find c so that f is continuous everywhere.


I'm assuming i would have to show lim x->c f(x)=f(c) or something?
I really dont know where to start with this question :S and it is not a homework question, just studying.

You are on the right track. Both pieces of the function are polynomials and are therefore continuous by themselves. The big deal is that the function changes types at x = 4. To have it be continuous at x = 4, you need these two pieces to "meet". That is, you need

lim f(x) = f(4).
x->4

Okay, so when x actually equals four, you have f(4) = c(4) + 20 = 20 + 4c. This value is also the limit as x->4 from the right. The left hand limit will involve the top part of f.

lim f(x) = 4² - c² = 16 - c²
x->4-

To get the limit to agree with f(4), solve the equation for c

20 + 4c = 16 - c².

I'll let you solve it. Turns out there is only one solution (the quadratic is a perfect square).

we did a question like this in calculus today. its a lot simpler than you think. first of all, you want the piecewise to be continuous, which means that the two pieces of the graph need to touch so that all points defined in the domain have a limit. basically, the end of the first piece needs to touch the end of the second piece. because the equations change at 4, both equations need to come to the same point when x=4 (or else you would have a jump and it wouldnt be continuous). since c is in both equations, you can plug 4 in for x and set the equations equal to eachother (since they would need the same y value).

4^2 - c^2 = c(4) +20

this is really simple math: c = 2

hope i helped :D

Since both "pieces" of function are polynomials, the only place where you might have problems with continuity is at x = 4

f(4) = c*4 + 20 = 4c + 20

limit[x→4⁺] = 4c + 20
limit[x->4⁻] = 4² - c² = 16 - c²

To ensure f(x) is continuous at x = 4, we need
limit[x→4⁺] = f(4) = limit[x->4⁻]
4c + 20 = 16 - c²
c² + 4c + 4 = 0
(c + 2)² = 0
c = -2

Function looks somewhat like this: http://www.wolframalpha.com/input/?i=y+%…
with the blue curve to left of 4 and pink line to right of 4
(just ignore both graphs above point of intersection)

You want to make sure that at x=4, x^2-c^2 = cx+20. (that is the two branches of the graph meet)
Solving this, you get c = -2.