Posted by **Anonymous** on Friday, October 29, 2010 at 6:25am.

Consider the function f(x) = x/(x-1)

Are there any turning points? Explain how this could help you graph f(x) for large values of x?

Ans: turning points is another word for checking the concavity, and therefore i find the second derivative and equate it to zero and see where it is concave up and down, using that information i can know when the graph is increasing or decreasing, with its end behaviors near the asymptote

I get the second derivative as f''(x) = 2/(x-1)^3. If f''(x) = 0 then there are no solutions, hence there are no turning points.

Am i correct? But how does that help me graph f(x) for large values.

- Math - Simple rational functions (check) -
**Reiny**, Friday, October 29, 2010 at 8:02am
"Turning points" is the same as maximum or minimum points, and we find them by setting the first derivative equal to zero.

Setting the second derivative equal to zero gives us

"points of inflection"

This page summarizes it for you

http://www.teacherschoice.com.au/Maths_Library/Calculus/stationary_points.htm

In most cases you can use common sense to find what happens to the graph for large values of x, by merely considering the numbers involved.

in your case consider a large x like x = 1000

the denominator is simply one less than the top, so we get 1000/999 which is barely above 1

the larger the x the closer we get to 1, but still above it

so for x ---> + infinity, f(x) = 1+

similarly for huge negative x's, e.g. x = -1000

we get -1000/-1001 which is barely below 1

so as x ---> - infinity , f(x) ---> 1-

- Math - Simple rational functions (check) -
**shila**, Friday, October 29, 2010 at 1:17pm
what is the anwser to this input output table here it is input(x)-2,-1,0,1,2,3,here is the output(y)-1,1,3,5,7,9,

- Math - Simple rational functions (check) -
**Valencia**, Thursday, February 9, 2012 at 5:16pm
3(x^2-x-6)/4(x^2-9) what is the domain?

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