I'm stuck on a problem.

For what value of the constant c is the function f(x) continuous on ( +/- infinity)

f(x)=x^{2}-c^{2}when x< 4

cx + 20 when x > or = to 4

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- May 19th 2013, 06:53 PMbaldysmcontinuity problem
I'm stuck on a problem.

For what value of the constant c is the function f(x) continuous on ( +/- infinity)

f(x)=x^{2}-c^{2}when x< 4

cx + 20 when x > or = to 4 - May 19th 2013, 07:04 PMProve ItRe: continuity problem
You have the function

$\displaystyle \displaystyle \begin{align*} f(x) = \begin{cases} x^2 - c^2 \textrm{ when } x < 4 \\ c\,x + 20 \textrm{ when } x \geq 4 \end{cases} \end{align*}$

Notice that the first part of this hybrid function is a quadratic, and the second part is a linear function. Both are polynomials, and so are continuous everywhere. Therefore, the ONLY point where the function may be discontinuous is when you have to go from one of these part functions to the next, so in this case, where $\displaystyle \displaystyle \begin{align*} x = 4 \end{align*}$.

In order for the function to be continuous there, it needs to be defined there (which it is), and it also needs to approach the same value from the left as from the right.

So what value will the function approach if you are approaching $\displaystyle \displaystyle \begin{align*} x = 4 \end{align*}$ from the left?

What value will the function approach if you approaching $\displaystyle \displaystyle \begin{align*} x = 4 \end{align*}$ from the right?

If these values have to be the same, what does that mean c has to be? - May 19th 2013, 08:09 PMbaldysmRe: continuity problem
I understand what your saying. I didn't ask the question well. I was just guessing at what C was and I'm looking for a mathematical way to solve it. I know the answer is C= -2.

I was thinking I need to set x^{2}-c^{2}= cx +20 and plug in 4 for the value of x, getting 16-c^{2}= 4c+20

Solving for C just gets ugly, uglier than this teacher would do on a test. So I am thinking I'm going down the wrong path, or am missing something. - May 19th 2013, 08:34 PMProve ItRe: continuity problem
That is exactly what you need to do. I don't see why you think solving for c would be ugly, it's a simple quadratic equation, solve as you normally would by making the equation equal 0 and either factorise, complete the square or use the Quadratic Formula...

$\displaystyle \displaystyle \begin{align*} 16 - c^2 &= 4c + 20 \\ 0 &= c^2 + 4c + 4 \\ 0 &= ( c + 2 ) ^2 \\ 0 &= c + 2 \\ -2 &= c \end{align*}$