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

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SUMMARY

The function f(x) is defined piecewise, with f(x) = cx² + 2x for x < 2 and f(x) = x³ - cx for x ≥ 2. To determine the value of the constant c that ensures continuity at x = 2, both one-sided limits must equal the function value at that point. The equation derived from setting the limits equal is c*4 + 4 = 8 - 2c, which simplifies to 6c = 4, yielding c = 2/3. Thus, the function is continuous for c = 2/3.

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Homework Statement


For what value of the constant c is the function f continuous on [tex](-\infty,\infty)[/tex]

[tex] f(x)=\left\{\begin{array}{cc}cx^2+2x,&\mbox{ if }<br /> x<2\\x^3-cx, & \mbox{ if } x\geq2\end{array}\right.[/tex]

Homework Equations


No idea :(


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 continuous at whatever values at x.. but not for a constant c, which is different. Pointers would be appreciated.
 
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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:
[tex]f(2)=2^{3}-c*2=8-2c[/tex]

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:
[tex]c*2^{2}+2*2=8-2c[/tex]
Solve this for c!
 
In order to be continuous, the limit must exist.
If it does then
[tex]\lim_{x\rightarrow 2}f(x)= \lim_{x\rightarrow 2^-} f(x)= \lim_{x\rightarrow 2^+}f(x)[/tex]

Now, what is
[tex]\lim_{x\rightarrow 2^-} f(x)= \lim_{x\rightarrow 2} cx^2+ 2x[/tex]
what is
[tex]\lim_{x\rightarrow 2^+} f(x)= \lim_{x\rightarrow 2} x^3- cx[/itex]<br /> <br /> Set them equal and solve for c.[/tex]
 

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