Use the given relationsfandgto find (a) f · g (b) f/g (f over g) (c) f∘g (d)g∘f

f(x)=Cos(x) g(x)=Csc(x)Tan(x)

can i see the steps on how to on these?

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- Jul 20th 2010, 03:23 PMRosieLairdf(x)=Cos(x) g(x)=Csc(x)Tan(x)
Use the given relations

*f*and*g*to find (a) f · g (b) f/g (f over g) (c) f∘g (d)g∘f

*f(x)=Cos(x) g(x)=Csc(x)Tan(x)*

*can i see the steps on how to on these?* - Jul 20th 2010, 03:36 PMtitos
first you have to fine the domain for f(x),g(x).Then give a name for each (example:h=f*g,d=f/g,l= f∘g,k=g∘f)then find the domain for the new functions(this very important to do),and for the end right the type for the new functions!!ok?

- Jul 20th 2010, 03:42 PMRosieLaird
how do you find the domain and range for them?

- Jul 20th 2010, 03:46 PMeumyang
I'll do (a)...

$\displaystyle \begin{aligned}

f \cdot g &= (cos \, x)(csc \, x)(tan \, x) \\

&= (cos \, x)\left( \frac{1}{sin \, x} \right) \left( \frac{sin \, x}{cos \, x} \right) \\

&= 1

\end{aligned}$

But this is not the same as your linear function y = 1. $\displaystyle f \cdot g$ is going to look like a horizontal line with some "holes". The domain of f(x) = cos(x) is all real numbers, but the domain of g(x) = csc(x)tan(x) is all reals except multiples of pi/2. So the domain of $\displaystyle f \cdot g$ is the intersection of the two domains, or all reals except multiples of pi/2. - Jul 20th 2010, 03:53 PMtitos
for the f(x) we see cos which is real in real numbers so domain of f is R

for g(x) it has y=css(x) wich means that the domain of that is R-{kπ},k in Z

and u=tan(x) has domain R-{kπ+π/2}k in Z

so domain for g(x) is R-{kπ+π/2,kπ},k in Z.ok now? - Jul 20th 2010, 11:33 PMMathelogician