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Math Help - Inverse Functions

  1. #1
    Junior Member Coco87's Avatar
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    Inverse Functions

    Hey, I've just started calc 2, I have two problems that are driving me crazy, and need some help

    The first one is probably extremely simple, but I've tried pretty much everything I can think of. The problem is that I need to find the inverse function of f(x):

    f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}

    I remember a similar problem involving conjugates, but I can't figure out how to apply it correctly (or I'm retarded ).

    The second one is an example our instructor gave us, but I can't figure out where she got the "0" for cos and sin -

    With this problem, we try to find the derivative of the inverse of f(x) where g(x) is the inverse:
    f(x)=x^3+3\sin{x}+2\cos{x}
    g(x)=f^{\neg1}(x)
    g^\prime(2)=?
    f^\prime(x)=3x^2+3\cos{x}-2\sin{x}
    2=x^3+3\sin{x}+2\cos{x}

    Now here's the part where she skipped a lot of stuff and literally put this on the board:

    \cos{0}=1
    \sin{0}=0

    g^\prime(2)=\frac{1}{3}

    Where did she get "0" from? I have no idea where that fits into the problem.

    Any help I can get would be awesome, even if it's just a hint.

    Thanks!
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  2. #2
    Moo
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    Hi !
    Quote Originally Posted by Coco87 View Post
    Hey, I've just started calc 2, I have two problems that are driving me crazy, and need some help

    The first one is probably extremely simple, but I've tried pretty much everything I can think of. The problem is that I need to find the inverse function of f(x):

    f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}

    I remember a similar problem involving conjugates, but I can't figure out how to apply it correctly (or I'm retarded ).
    If you want to apply conjugates...

    The conjugate of a+b \sqrt{c} is a-b \sqrt{c}. The product of the two is (a+b \sqrt{c})(a-b \sqrt{c})=a^2-b^2c by the identity (x-y)(x+y)=x^2-y^2.

    f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}} multiply by \frac{1-\sqrt{x}}{1-\sqrt{x}}=1 :

    f(x)=y=\frac{(1-\sqrt{x})^2}{1-x}

    But I think it won't help ^^'
    ---------------------------
    y=\frac{1-\sqrt{x}}{1+\sqrt{x}} \implies y+y \sqrt{x}=1-\sqrt{x}

    \sqrt{x}(y+1)=1-y

    \sqrt{x}=\frac{1-y}{1+y} \implies x=\left(\frac{1-y}{1+y}\right)^2=\boxed{f^{-1}(y)}


    The second one is an example our instructor gave us, but I can't figure out where she got the "0" for cos and sin -

    With this problem, we try to find the derivative of the inverse of f(x) where g(x) is the inverse:
    f(x)=x^3+3\sin{x}+2\cos{x}
    g(x)=f^{\neg1}(x)
    g^\prime(2)=?
    f^\prime(x)=3x^2+3\cos{x}-2\sin{x}
    2=x^3+3\sin{x}+2\cos{x}

    Now here's the part where she skipped a lot of stuff (SiMoon says : I don't think she did, you'll see below ) and literally put this on the board:

    \cos{0}=1
    \sin{0}=0

    g^\prime(2)=\frac{1}{3}

    Where did she get "0" from? I have no idea where that fits into the problem.

    Any help I can get would be awesome, even if it's just a hint.

    Thanks!
    Ok

    x=f(f^{-1}(x))=f(g(x))

    Let's apply chain rule :

    1=g'(x)\cdot f'(g(x))

    Therefore g'(2)=\frac{1}{f'(g(2))}

    We have to calculate g(2), that is to say x such that 2=x^3+3\sin{x}+2\cos{x}, because g(2)=f^{-1}(2) \implies 2=f(x)

    What your teacher did was to note that there is a 2 in the LHS. There's a coefficient 2 in the RHS. And magically, if you put x=0, it works :
    2=0^3+3\sin(0)+2\cos(0)=2
    Hence g(2)=0.

    Therefore g'(2)=\frac{1}{f'(g(2))}=\frac{1}{f'(0)}=\frac 13

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  3. #3
    o_O
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    If you need to be picky, note that the range of your original function is -1 < y \leq 1 so the domain of your inverse function will be -1 < x \leq 1.
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  4. #4
    Junior Member Coco87's Avatar
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    Quote Originally Posted by Moo View Post
    Hi !

    If you want to apply conjugates...

    The conjugate of a+b \sqrt{c} is a-b \sqrt{c}. The product of the two is (a+b \sqrt{c})(a-b \sqrt{c})=a^2-b^2c by the identity (x-y)(x+y)=x^2-y^2.

    f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}} multiply by \frac{1-\sqrt{x}}{1-\sqrt{x}}=1 :

    f(x)=y=\frac{(1-\sqrt{x})^2}{1-x}

    But I think it won't help ^^'
    ---------------------------
    y=\frac{1-\sqrt{x}}{1+\sqrt{x}} \implies y+y \sqrt{x}=1-\sqrt{x}

    \sqrt{x}(y+1)=1-y

    \sqrt{x}=\frac{1-y}{1+y} \implies x=\left(\frac{1-y}{1+y}\right)^2=\boxed{f^{-1}(y)}
    Haha, I guess I was way off on that one Seems to be my problem, not good at finding answers like that.

    Quote Originally Posted by Moo View Post
    Ok



    Let's apply chain rule :



    Therefore

    We have to calculate g(2), that is to say x such that , because

    What your teacher did was to note that there is a 2 in the LHS. There's a coefficient 2 in the RHS. And magically, if you put , it works :

    Hence .

    Therefore

    Wow, did I completely over analyze that I understand now.

    Thanks!
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