Finding the max/min values of this function?

1. Jan 7, 2017

Vitani11

1. The problem statement, all variables and given/known data
Analyze the function f(x) = [(x2-7x+12)/(x-1)]e-x/2. Identify the zeroes, local maxima, and minima (approximate locations), as well as points where the function "blows up", give asymptotic values (for x→±∞), and sketch the function.

2. Relevant equations
f(x) = [(x2-7x+12)/(x-1)]e-x/2
f'(x) = (e-x/2(12-7x+x2)+(-1+x)[e-x/2(-7+2x)-((e-x/2)/2)(12-7x+x2)))
3. The attempt at a solution
Above is the derivative. I can't just solve this thing for x, can I? It's hellish and involves logs and there is a high chance that I'll mess up. However the professor said approximate. So how do I do that? I was never formally taught how to approximate a functions local max or min. Do I binomial expand each term in the derivative after simplification of it and then find the zeros?

What about the asymptotes? Can't just do long division to find it because there is an e-x/2 in there...

Last edited: Jan 7, 2017
2. Jan 7, 2017

Staff: Mentor

By asymptotes, I think they mean for you to find places where f is undefined. Think about where the denominator is zero.

For f', combine the fractions you get into a single fraction, all multiplied by e-x/2. Since e-x/2 is never zero, see if you can at least approximate where the numerator of the fraction is zero.

3. Jan 7, 2017

Staff: Mentor

You can look for zeros exactly, and you can identify where the function "blows up", and how the function looks for very large and very small x. Combine that to make a sketch of the function. You can also use a computer (e.g. calculate the value of the function for various values of x).

The derivative looks wrong, and it won't help much. Edit: Turns out the correct derivative does help.

Last edited: Jan 8, 2017
4. Jan 7, 2017

Vitani11

The derivative is all of that over (x-1)2. Sorry there was so much in it to write that I forgot a portion of it. I know this is correct now because of Wolfram Alpha. The zeroes I have no problem with. I have to do all of the things listed though, I can't just sketch the function using other means unfortunately. How do I go about finding the max and mins? What approximation technique should be used? Same with asymptotes.

5. Jan 7, 2017

Staff: Mentor

Write the derivative as $f'(x) = \frac{\text{stuff}}{(x - 1)^2} e^{-x/2}$. f' will be zero at the places where "stuff" is zero. That's what I was talking about when I said to approximate where the numerator is zero.
You could also look at $g(x) = \frac{(x - 4)(x - 3)}{x - 1}$, (ignoring the exponential term temporarily). Multiplying by $e^{-x/2}$ will have the effect of "scrunching" the graph of g down as you move to the right. Just a thought...

6. Jan 7, 2017

Thank you

7. Jan 7, 2017

Ray Vickson

In fact, "stuff" = a cubic polynomial, so is relatively easy to deal with. Its roots are not "nice", however, but there are three of them, all real.

8. Jan 9, 2017

Vitani11

Okay... so I've gotten down to that part. I am told that my best bet to approximate the roots of this cubic is to use newtons method since it is not factorable. But without cheating (by graphing, wolfram, etc.) how am I to come up with an initial value to do the approximation? What about a linear approximation?

9. Jan 9, 2017

Staff: Mentor

You can calculate a few example values. Every time the sign changes there has to be a root in between.

10. Jan 9, 2017

Vitani11

Got it! Wooo! Last question, what does it mean to find a point when a function "blows up?" Just where the limit approaches infinity around the asymptote, so the x coordinate of where the asymptote is located?

11. Jan 9, 2017

Staff: Mentor

Yes. The function "blows up" because one factor in the denominator is getting close to zero.

12. Jan 9, 2017

Ray Vickson

You can apply Descarte's Rule of signs to your polynomial $p(x)$ conclude that for positive $x$ there are either 3 roots or 1 root, and for negative $x$ there are no roots. In fact, since $p(0) = -2$ and $p(1) = 12,$ the point $x = 0$ must be pretty close to a root; thus, starting Newton from 0 seems a good choice.