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Math Help - Trig functions

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    Trig functions

    Trig functions-math2.jpg how do you find (a) f g (b) f/g (f over g) (c) f∘g (d)g∘f
    It's for the problem above the bold lettering. Just above it.It has the 5 root and 3 root in it

    Also I am stumped on how to find the domain and range for it as well
    Last edited by RosieLaird; July 23rd 2010 at 12:31 PM.
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    Member Mathelogician's Avatar
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    To find (a) and (b), use the difinitions of cscx=1/sinx and tanx=sinx/cosx.
    For (c) and (d) first clarify which f and g do you mean and then use definition of composition of functions (f∘g)(x)=f(g(x)) and subsituate g(x) instead of x in f. Similar for g∘f.
    Domain and range for which ones?
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    the domain and range for all of them
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    Senior Member eumyang's Avatar
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    Quote Originally Posted by RosieLaird View Post
    how do you find (a) f g (b) f/g (f over g) (c) f∘g (d)g∘f
    It's for the problem above the bold lettering. Just above it.It has the 5 root and 3 root in it

    Also I am stumped on how to find the domain and range for it as well
    f(x) = \dfrac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}}
    g(x) = \dfrac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}}

    For (a), just multiply the two functions together:
    (f \cdot g)(x) = \left( \dfrac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}} \right) \left( \dfrac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}} \right)

    For (b), divide the two functions together (which is the same as multiplying by the reciprocal):
    (f/g)(x) = \left( \dfrac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}}  \right) \left( \dfrac{1 - \sqrt[3]{x + 1}}{2x - \sqrt{x}} \right)

    For (c), in f(x), plug in g(x) for x:
    (f \circ g)(x) = \dfrac{6 \left( \sqrt{\frac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}}} \right) + \left( \frac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}} \right)^2}{\sqrt[5]{\left( \frac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}} \right) - 2}}

    For (d), in g(x), plug in f(x) for x:
    (g \circ f)(x)  = \dfrac{2 \left( \frac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}} \right) - \sqrt{\frac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}}}}{1 - \sqrt[3]{\left( \frac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}} \right) + 1}}

    I hope that you don't have to simplify these.
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