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Math Help - comparing graphs of functions

  1. #1
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    comparing graphs of functions

    Compare the functions f(x) = x7 and g(x) = 7x by graphing both functions in several viewing rectangles. (a) Find all points of intersection of the graphs correct to one decimal place.
    ( 1, 2) (smaller x value)
    ( 3, 4) (larger x value)


    I cant seem to find the points when i put it in my calculator. any help?
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  2. #2
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    Quote Originally Posted by kyleu03 View Post
    Compare the functions f(x) = x7 and g(x) = 7x by graphing both functions in several viewing rectangles. (a) Find all points of intersection of the graphs correct to one decimal place.
    ( 1, 2) (smaller x value)
    ( 3, 4) (larger x value)


    I cant seem to find the points when i put it in my calculator. any help?
    what is the significance of the red 7 ?

    is it an exponent in f(x) ?

    is
    g(x) = 7x or 7^x ?

    use a caret (^) symbol to denote exponents ... i.e. x^7 is x to the 7th power.
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    it is just the number, it is red because its changes. idk how to do it
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    Quote Originally Posted by kyleu03 View Post
    it is just the number, it is red because its changes. idk how to do it
    This does not answer the questions raised and will just delay you getting a helpful reply. Re-post the equations. Use the conventional formatting for exponents (that is, powers).

    Note: x^7 means x^7, 7^x means 7^x,
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    Quote Originally Posted by kyleu03 View Post
    it is just the number, it is red because its changes. idk how to do it
    x7 makes no sense. what is f(x) ?
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    ohh im very sorry i didnt see it posted wrong. it is

    f(x) = x^7
    g(x) = 7^x
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  7. #7
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    Quote Originally Posted by kyleu03 View Post
    ohh im very sorry i didnt see it posted wrong. it is

    f(x) = x^7
    g(x) = 7^x
    one solution should be obvious ...

    graph y = x^7 - 7^x and look for the the other zero ... easier to find.
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    "graph and look for the the other zero " does not find the points of intersection for me
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    Quote Originally Posted by kyleu03 View Post
    "graph and look for the the other zero " does not find the points of intersection for me
    An obvious solution is a whole number that lies somewhere between 5 and 8 ....

    The other solution solution cannot be found exactly. A decimal approximation lies somewhere between 1 and 2. You need to use your graph (or graphs) to estimate the value correct to 1 decimal place.

    As to why you can't find the intersection points when you put it into your calculator, you should check:

    1. that you've entered the correct equations.

    2. you're using an appropriate window (check that ymin and ymax of the window are OK).
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    I was able to find the smaller, but not larger. any help?


    Compare the functions f(x) = x7 and g(x) = 7x by graphing both functions in several viewing rectangles.
    (a) Find all points of intersection of the graphs correct to one decimal place.
    ( 1, 2) (smaller x value)
    ( 3, 4) (larger x value)

    (b) Which function grows more rapidly when x is large? 5
    f(x)
    g(x)







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  11. #11
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    Quote Originally Posted by mr fantastic View Post
    An obvious solution is a whole number that lies somewhere between 5 and 8 ....

    The other solution solution cannot be found exactly. A decimal approximation lies somewhere between 1 and 2. You need to use your graph (or graphs) to estimate the value correct to 1 decimal place.

    [snip]
    Quote Originally Posted by kyleu03 View Post
    I was able to find the smaller, but not larger. any help?
    [snip]
    There aren't many whole numbers between 5 and 7 that require testing ....
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  12. #12
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    are you saying (0,7)
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  13. #13
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    Quote Originally Posted by kyleu03 View Post
    are you saying (0,7)
    I am saying that you should check what happens when x = 6 and x = 7.
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  14. #14
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    thats wrong blong
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  15. #15
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    Quote Originally Posted by kyleu03 View Post
    thats wrong blong
    For crying out loud! f(x) = x^7 and g(x) = 7^x have an intersection point at x = 7. The value of each function at x = 7 is 7^7. Given the previous replies, this should not have needed spelling out.
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