1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Combining Functions Question

  1. May 27, 2009 #1
    Question:
    Consider the graph of f(x)=g(x)/h(x) where g(x) = 2^x and h(x)=x

    a) Explain why it makes sense that the graph is in only the 1st and 3rd quadrants
    b) Explain the behaviour of the curve in the first quadrant by making reference to the original functions.


    I am sooo bad at these communication questions and I desperately need help!
     
  2. jcsd
  3. May 27, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    For the quadrants, consider the signs of the x values with the corresponding sign of the y values.

    For example, in the first quadrant, is x-positive or negative? and is y positive or negative?
     
  4. May 27, 2009 #3
    In QI, the x and y values of g(x) and h(x) in the first quadrant are positive.
    The x and y values of h(x) in QIII are all negative, but g(x) never goes into QIII because of the asymptote. How do I explain that? lol.
     
  5. May 27, 2009 #4

    rock.freak667

    User Avatar
    Homework Helper

    For all values of x, 2x is postive. So for 2x/x and x is negative you will have positive/negative which gives a negative number, which would correspond to the sign of the y values in quadrant 3
     
  6. May 27, 2009 #5
    Thanks a lot, that makes a lot more sense. Can you help me with the end behaviours questions?
     
  7. May 27, 2009 #6

    rock.freak667

    User Avatar
    Homework Helper

    In each quadrant, describe what happens as x becomes increasingly positive or negative as the case may be. Explain if there would be any significant features at any spefic value(s) of x and such.
     
  8. May 27, 2009 #7
    Well, in QI, as x gets bigger, 2^x quickly gets closer toward infiniti. Similarly, f(x)=x also gets closer to infinite, but at a slower rate. In QIII, f(x)=x goes to negative infiniti.
     
  9. May 27, 2009 #8

    rock.freak667

    User Avatar
    Homework Helper

    Right so f(x) will tend to infinity.

    so as x goes to negative infinity 2x will tend to what value?
     
  10. May 27, 2009 #9
    Zero, right?
     
  11. May 27, 2009 #10

    rock.freak667

    User Avatar
    Homework Helper

    yes. So would 2x reach zero before x reaches negative infinity or does x reach before 2x?
     
  12. May 27, 2009 #11
    2^x would reach zero before x reaches negative infiniti, correct?
     
  13. May 27, 2009 #12

    rock.freak667

    User Avatar
    Homework Helper

    good good....anthing special happens when x goes to zero?
     
  14. May 27, 2009 #13
    are you refering to the asymptote?
     
  15. May 27, 2009 #14
    Although it's good practice (and fun too) to consider both quadrants, just thought I'd point out that the problem only needs an answer for first quadrant for part b).
     
  16. May 27, 2009 #15
    Yeah, sorry I got side tracked. I'm looking at the two parent functions and I can't figure out how to explain the new end behaviours of the curve in QI. :S
     
  17. May 27, 2009 #16
    It appears you and rock did a pretty good job of this already. You just need to put it all together and summarize. You found the asymptote as x goes to zero by considering what g(x) and h(x) does, and you described the fact the the function blows up at x towards infinity by considering how fast the numerator g(x) blows up compared to h(x). Just need to put it down on paper.

    Now, some other things you might want to write. Does the Quadrant I part of the function stay within Quadrant I? Or does it cross the x-axis at some point? How do you know? (You already know it doesn't cross the y-axis because it blows up there.)

    If it goes to positive infinity at x=0 and blows up at x=Inf, what does that imply that the function has at least one of in the middle? (Hint: think about minimums and maximums in the curve)
     
  18. May 27, 2009 #17
    Yeah, that part of the curve is all in QI because you can clearly see that both the end behaviours are going to positive infiniti. As for the second part, does that imply that there's at least one asymptote?
     
  19. May 27, 2009 #18
    That's not good enough to say that. It could fall from positive infinity at x=0, fall past the x-axis line to negative y, then turn around and head back up, cross the x-axis line back to positive and then go to positive infinity at x=infinity. As Rock pointed out, one way to know that doesn't happen is that g(x) and h(x) both stay positive so the result has to stay positive and never goes negative.

    However, I'm saying there is a second feature of the graph that you could also talk about. What must happen to the numerator, g(x), if the function were to cross the x-axis? What happens to g(x) at the moment the function just crosses. However, Rock's answer showing that it is always positive is sufficient and you could ignore this part if you want.

    No, that's not what I was fishing for. I was asking about function minimums and maximums. I don't know if you've gone that far in your course yet. It's probably not necessary to consider it.
     
  20. May 27, 2009 #19
    Well, isn't the reason why it shoots up to positive inifiniti near x=0 because it's 2^x divided by a really small X value from the f(x)=x curve? Like, when you divide a a number by a tiny number, it gives you a huge number, correct? And as for the other side of the curve, the 2^x function is still going up faster than f(x)=x which is why it continues in an exponential-esque shape. Correct?
     
  21. May 27, 2009 #20
    Yes.

    But your stuck on looking at the ends. That may be fine for your problem. What I was trying to get you to consider is what happens in the middle.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Combining Functions Question
  1. Combination question (Replies: 1)

  2. Combining Functions (Replies: 1)

Loading...