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Math Help - even/odd graph

  1. #1
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    Post even/odd graph

    Are the graphs of

    \sqrt{x}

    2^x

    log_2x

    even or odd? I don't see any symmetry with respect to y axis or origin. Thanks!
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  2. #2
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    Quote Originally Posted by live_laugh_luv27 View Post
    Are the graphs of

    \sqrt{x}

    2^x

    log_2x

    even or odd? I don't see any symmetry with respect to y axis or origin. Thanks!
    A function f(x) is even if f(-x) = f(x)

    A function f(x) is odd if f(-x) = -f(x)

    can you continue?
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  3. #3
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    Post

    so,

    even

    odd

    even

    ?
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  4. #4
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    I'm still not exactly sure how to set that up, but have another ?...even if a function does not have symmetry in respect to the y-axis, can it still be even? similarly, even if the function does not have symmetry in respect to the origin, can it still be odd?
    thanks!
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  5. #5
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    Quote Originally Posted by live_laugh_luv27 View Post
    so,

    even

    odd

    even
    What does any of the above have to do with the basic definitions of odd & even?
    \text{Odd functions are such that }f(-x)=-f(x)

    \text{Even functions are such that }f(-x)=f(x)
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by live_laugh_luv27 View Post
    I'm still not exactly sure how to set that up, but have another ?...even if a function does not have symmetry in respect to the y-axis, can it still be even? similarly, even if the function does not have symmetry in respect to the origin, can it still be odd?
    thanks!
    no, even and odd follow the exact definitions i gave you. you don't even have to look at the graphs.

    example, state whether (a) f(x) = x^2, (b) f(x) = \sin x and (c) f(x) = \frac {x^2}{x^3 + 1}

    (a) f(x) = x^2 is even, since

    f(-x) = (-x)^2 = x^2 = f(x)

    (b) f(x) = \sin x is odd, since

    f(-x) = \sin (-x) = - \sin x = - f(x)

    (c) f(x) = \frac {x^2}{x^3 + 1} is neither even nor odd, since

    f(-x) = \frac {(-x)^2}{(-x)^3 + 1} = \frac {x^2}{1 - x^3} \ne f(x) \text{ or } -f(x)


    notice that i didn't even draw any graphs. now, try again
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  7. #7
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    Quote Originally Posted by Plato View Post
    What does any of the above have to do with the basic definitions of odd & even?
    \text{Odd functions are such that }f(-x)=-f(x)

    \text{Even functions are such that }f(-x)=f(x)
    so these answers are incorrect?
    - even

    - odd

    - even
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by live_laugh_luv27 View Post
    so these answers are incorrect?
    - even

    - odd

    - even
    yes, they are wrong. did you try what i said?
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  9. #9
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    = \sqrt{-x} = odd

    = 2^{-x} = odd?

    = \frac{ln(-x )}{ln2} ?
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    Quote Originally Posted by live_laugh_luv27 View Post
    = \sqrt{-x} = odd
    ?! are you telling me that, for example, \sqrt 2 = \sqrt {-2}? and if so, does that comply with the definition of an odd function as defined in my and Plato's posts?

    = 2^{-x} = odd?
    interesting. say x = 1, is it true that 2^1 = 2^{-1} or -2^1 = 2^{-1} ?

    = \frac{ln(-x )}{ln2} ?
    tell me, what's \log_2 (-5), say
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  11. #11
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    = = -\sqrt{x} = -f(x) ?

    is not true

    is \frac{ln(-5)}{ln(2)} ?
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  12. #12
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    Quote Originally Posted by live_laugh_luv27 View Post
    = = -\sqrt{x} = -f(x) ?

    is not true

    is \frac{ln(-5)}{ln(2)} ?
    ok, as far as real numbers are concerned, the square root function and the logarithm function are not defined for negative real numbers.... you should know this. please look this up and make sure you get it

    and clearly 2^1 \ne 2^{-1} nor does -2^1 = 2^{-1} so that f(x) = 2^x is neither even nor odd.
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