inverses

[Guest]

the problem is i just don't know. i seem to do ok until i get sq roots

verify that f(x) and f-1 are inverses of each other

f(x) = sq root x-2

 

[stapel]

verify that f(x) and f-1 are inverses of each other

The method for verification should have been mentioned in class and/or in the text: Compose the two functions, and confirm that, upon simplification, the result is just "x".

f(x) = sq root x-2

Your formatting is ambiguous. Do you mean "sqrt[x] - 2", "sqrt[x - 2]", or something else? And what is f^-1?

Thank you.

Eliz.

 

[Gene]

If you want a superscript you can use
^-1 to show
f^-1 and
f(x)=sqrt(x^-2)

BTW Replacing sup with sub gives a subscript.

 

[Guest]

i would like to thank evry-one for there imput. mostly directions on how to put in sub scrips thank you.

yes the class provided the method. i can not get the other function in order to compose them
pardon my lack of skills here.

i need help with finding the inverse of f(x)=sqrt(x-2)
x-2 is all under the radical

? is verify that f(x) and f ^-1 are inverses of each other.

 

[pka]

When one is working with even roots care must be exercised
If \(\displaystyle f(x) = \sqrt {x - 2} ,\quad x \ge 2\) note the domain. Moreover, the range is \(\displaystyle f(x) \ge 0\).

Consider \(\displaystyle g(x) = x^2 + 2\).

Thus if \(\displaystyle f(x) \ge 0\) then \(\displaystyle \begin{eqnarray}
g(f(x)) & = & \left( {\sqrt {x - 2} } \right)^2 + 2 = x \\
f(g(x)) & = & \sqrt {\left( {x^2 + 2} \right) - 2} = \left| x \right| \\
\end{array}\).

Therefore we must say that the domain of g is \(\displaystyle x \ge 0\).

Having done that then \(\displaystyle g(x) = f^{ - 1} (x)\).