# For what value of the constant c is f(x) continuous?

1. Jul 28, 2008

### illjazz

1. The problem statement, all variables and given/known data
For what value of the constant c is the function f continuous on $$(-\infty,\infty)$$

$$f(x)=\left\{\begin{array}{cc}cx^2+2x,&\mbox{ if } x<2\\x^3-cx, & \mbox{ if } x\geq2\end{array}\right.$$

2. Relevant equations
No idea :(

3. The attempt at a solution
I tried looking at the examples preceding this section's problem section but could not find anything quite resembling this. There are examples for finding where f would be continous at whatever values at x.. but not for a constant c, which is different. Pointers would be appreciated.

2. Jul 28, 2008

### arildno

Well, if a function is continuous at a point (here, evidently, the only point of problem is x=2!), then both its one-sided limits must equal the function value AT x=0.

For a given c, the function value f(2) is given by the lower expression:
$$f(2)=2^{3}-c*2=8-2c$$

Now, within its domain, the upper expression is just a polynomial in x, i.e continuous.

That means that f(2) must equal whatever value the upper expression gains AT 2.
This gives you the equation for c:
$$c*2^{2}+2*2=8-2c$$
Solve this for c!

3. Jul 28, 2008

### HallsofIvy

Staff Emeritus
In order to be continuous, the limit must exist.
If it does then
$$\lim_{x\rightarrow 2}f(x)= \lim_{x\rightarrow 2^-} f(x)= \lim_{x\rightarrow 2^+}f(x)$$

Now, what is
$$\lim_{x\rightarrow 2^-} f(x)= \lim_{x\rightarrow 2} cx^2+ 2x$$
what is
[tex]\lim_{x\rightarrow 2^+} f(x)= \lim_{x\rightarrow 2} x^3- cx[/itex]

Set them equal and solve for c.