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Is my teacher wrong?
chz3369 #1 Sep 9, 2022, 11:29 PM
So I encountered a problem in Precalc, and it is to find the range of a rational function:$$f(x)=\frac{x-4}{x^2-x-6}.$$What my teacher said was the range was all real numbers, because there are no holes (numerator and denominator doesn't cancel), and the function crosses the horizontal asymptote $y=0$ at $x=4$, so she concluded that the range is all real numbers, $(-\inf, \inf)$. However, while trying to find the range by myself with a different method, I found some problems with this.
My method:
To test the range, I came up setting the polynomial to an arbitrary real number $r$, and solving to find any values for which $r$ can't be. So, I had$$\frac{x-4}{x^2-x-6}=r \rightarrow rx^2-rx-6r=x-4$$$$rx^2-(r+1)x-(6r-4)=0.$$
When a value of $r$ is not possible, it necessarily means that the left hand side (which is the rational function) will produce an imaginary result or a contradictory statement (like $1=0$). Because the $x^2$ term will always exist unless $r=0$ (which we find to be possible with $x=4$), this makes the imaginary result the only possibility to test for. So, to test for possible imaginary values, we set the discriminant of our polynomial with $r$ to be less than $0$:$$(r+1)^2-4\cdot r\cdot -(6r-4)<0.$$
This simplifies to$$25r^2-14r+1<0,$$which has solutions$$\frac{7-2\sqrt6}{25}<r<\frac{7+2\sqrt6}{25}.$$Then this is the range where $r$ values will produce an imaginary result on the LHS of the original polynomial $\dfrac{x-4}{x^2-x-6}$.
So who is right? Is there a problem with my method? All explanations are much appreciated.
Funny.Math #2 Sep 10, 2022, 1:11 AM
Your idea is good, but you made a mistake. The equality $~rx^2-(r+1)x-(6r-4)=0~$ it is true for all REAL $x$, so we must have $~{{(r+1)}^{2}}+4\cdot r\cdot (6r-4)\ge 0~$, and just solve this inequality.
To check that everything is fine, use the derivative of the function and the monotonity, if it is consistent with the above, then everything is fine.
greenturtle3141 #3 Sep 10, 2022, 1:35 AM
I disagree with the above. @OP, your method is completely correct, and hence so is your answer.
You can verify your work visually. Try graphing $f(x)$ in the Desmos Graphing Calculator, and also graph $~y = \dfrac{7+2\sqrt{6}}{25}~$ and $~y = \dfrac{7-2\sqrt{6}}{25}~$. What you will see will confirm your work here.
Funny.Math #5 Sep 10, 2022, 2:10 AM
That's right, you're right, I didn't even say that for all real r give a rel x as a solution. Those r numbers are the good ones, which are the solutions to the inequality I described: ${{(r+1)}^{2}}+4\cdot r\cdot (6r-4)\ge 0$.
By the way, check the image here, it matches the solution you get from the inequality!
https://www.wolframalpha.com/input?i=plot+%28x-1%29%2F%28x%5E2-x-6%29
Funny.Math #6 Sep 10, 2022, 2:11 AM
Just see this:
https://www.wolframalpha.com/input?i=plot+%28x-1%29%2F%28x%5E2-x-6%29
The image it is not in concordance with this: $\dfrac{7-2\sqrt6}{25}<r<\dfrac{7+2\sqrt6}{25}.$
alexheinis #12 Sep 10, 2022, 8:29 PM
The OP's solution is fine and I agree with Greenturtle that the remark in #2 is wrong. The formula in the Wolfram Code of FunnyMath in #6 is wrong since the function is $\displaystyle f(x)={{x-4}\over {x^2-x-6}}$ and not $~\displaystyle \color{red}{f(x)={{x-1}\over {x^2-x-6}}}$. Hence #6 can be discarded. The post in #9 is a direct copy op OP's solution, the final answer is wrong since $-2,3$ are contained in the range. The person in #9 confuses domain and range.
Funny.Math #13 Sep 10, 2022, 9:48 PM
Yes, it was a mistake, but it doesn't change the essence!
The correct WA link:
https://www.wolframalpha.com/input?i=plot+%28x-4%29%2F%28x%5E2-x-6%29
This is in concordance with the $\dfrac{7-2\sqrt6}{25}<r<\dfrac{7+2\sqrt6}{25}.$
alexheinis #14 Sep 10, 2022, 11:03 PM
It does change the essence because the graph with x-1 has full range
Correcting people on the basis of false information does not do your credibility any good |
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发表于 2023-10-28 00:06
