Making a piecewise function continuous everywhere

[ninty45]

Homework Statement

Given the piecewise function

$f(x) = \left\{ \begin{array}{lr} \frac{(2-x)^2-p}{x} &: x < q\\ r(x+6) &: q \leq x <2 \\ x^3-p &: x \geq 2 \end{array} \right.$

Find the values of p,q,r such that f(x) is continuous everywhere and f(2) = p

The Attempt at a Solution

Since f(2) = p,
$f(2) = (2)^3 - p = p\\ 8-p=p\\ p=4$

If f(x) is continuous everywhere, then

$\lim_{x \to 2} f(x) = f(2)$

But f(2)=p, and r(x+6) is always smooth, hence

$r(2+6) = 4 \\ r = \frac{1}{4}$

My difficulty is with q. Again from (assumed) continuity, at x=q,

$\lim_{x \to q} f(x) = f(q) \\ \lim_{x \to q} \frac{(2-x)^2-4}{x} = \frac{1}{2}(q+6) \\ \lim_{x \to q} x-4 = \frac{1}{2}(q+6) \\ 2q-8 = q+6 \\ q=14$

but from the original function, q <2. The answer given is supposedly q=0, but I do not see then how the function could be continuous.

Are my calculations wrong? I tried assuming that q =0, hence instead of factorising the limit I used l'hopital but no dice.

I am including the original picture as a comparison.

 

[PeroK]

$r(2+6) = 4 \\ r = \frac{1}{4}$

There's a bit of a error there, I'm sorry to say.

 

[ninty45]

My mistake. I wrote that incorrectly, it should be $r=\frac{1}{2}$.

However in the following working, I indeed used $r=\frac{1}{2}$.

 

[HallsofIvy]

You are correct. In order that f(2)= 2, p must be 4 and in order that the function be continuous at x= 2, r must be 1/2 (you say "1/4" but I presume that is a typo since you later use "1/2"). But in that case, there is NO value of q, less than 2, such that $[(2- q)^2- 4]/q$ is equal to (1/2)(q+ 6) so there is no value of q, less than 2, such that f(x) is continuous at x= q.

 

[PeroK]

My mistake. I wrote that incorrectly, it should be $r=\frac{1}{2}$.

However in the following working, I indeed used $r=\frac{1}{2}$.

The question looks wrong. Also, it doesn't cover the case where x = 0, so I assume there is a typo somewhere.

 

[ninty45]

Thanks for the replies. I'll just chalk this one up as one of those incorrect questions.