e^(x+1) >/= (2-x)/(x-1)
My textbook gives these instructions :
Sketch the 2 equations on the same diagram and find x-coordinate of all intersections.
Since e^(x+1) is greater or equals to (2-x)/(x-1),
write the range of x such that e^(x+1) is higher than (2-x)/(x-1).
I know how to whole thing goes and all but im not quite sure how to determine which range of numbers to draw out from the graph.
Please help? thank you very much!
Thanks for the help so far!
Yes. this is exactly where im stuck!
I do not know how to find the intervals.
My teacher taught us the method of shading. But how do i know where to shade?
Thank you very much! I hope im not confusing you.
You could shade along the x-axis for all those points in which the exponential function has a greater value than the rational function. Can you state what you think the intervals are?
You have the right idea.
x < 1 is correct, and I would, for the other interval, round to 1.11 instead and use a weak inequality (≥) since both functions are defined there and the original inequality is weak.
Are you expected to give the solution in interval or set-builder notation?
Here's another way to solve inequalities graphically; I think it's worthwhile adding to your arsenal. Transform the given inequality to one with 0 on one side. Let f(x) be the non-zero side. For example,
Now list in increasing order all x's with f(x)=0 or f(x) undefined. This gives you a set of open intervals. Choose one point c in each interval and find the sign of f(c). Then f(x) has the same sign as f(c) for all x's in the open interval containing c.
The above works for "nice" functions; if you go on and take calculus, you'll learn what I mean by "nice" functions. This is not a graphical solution, but it does provide a check on your reading of the graph of f(x). The attachment shows this method of attack: