algebra

[Elaine]

I don't know how to reply to thank you all and answer questions when your helping me through some problems and asking me questions. Thank you for the math advice..and way to write it -it's harder writing math problems on line until you know how to do it.

ok , heres a couple more to check please for me....

f(x)and g(x)such that h(x)=(f og)(x), if h(x)= square root 1+1/x

is the solution ..f(x)=sq.rt (x+1),g(x)=1/x

f(x)=x^2-7 is not a one to one function. Is that an accurate statement?

Finally,

An equation for the inverse of the relation y=4x-6

Is it y=(x-4)/6

thanks so much..it's nice to know if I'm doing these problems right b/c I have about 15 of each to do.... Thanks!!!

 

[wjm11]

An equation for the inverse of the relation y=4x-6

Is it y=(x-4)/6

Hi, Elaine,

To find the inverse, start by switching x and y, then solve for y:

Y = 4x – 6 becomes (new, inverse function)
X = 4y – 6 (add 6 to each side)
X + 6 = 4y (divide each side by 4)
Y = (x + 6)/4 = (1/4)x + 3/2

Note: if you plot a function and its inverse, you’ll find that they are symmetric with respect to the line y = x. That’s one way to check your work. Additionally (and more rigorously), if f(x) and g(x) are inverses, then f(g(x)) = g(f(x)) = x.

Hope that helps.

 

[stapel]

Your post is somewhat ambiguous, so I will be restating the questions according to my understanding of them. Please reply with corrections if my guesses are wrong. Thank you.

1) Find f(x) and g(x) such that h(x) = f(g(x)) = sqrt(1 + 1/x).

Work from the outside in. What is the first operation that you get to in "sqrt(1 + 1/x)"? That would be what h(x) did to g(x). What is inside that? That would be what g(x) did to x.

Your answer is one valid solution:

. . . . .f(x) = sqrt(x + 1), g(x) = 1/x
. . . . .f(g(x)) = f(1/x) = sqrt([1/x] + 1]

Another solution might be:

. . . . .f(x) = sqrt(x), g(x) = 1 + 1/x

There are, no doubt, many other answers, but these are probably the simplest.

2) Find the inverse of the relation y = 4x - 6.

To check inverses, compose:

. . . . .f(x) = 4x - 6
. . . . .f^-1(x) = (x - 4)/6
. . . . .f(f^-1(x)) = f([(x - 4)/6])
. . . . .= 4[(x - 4)/6] - 6 = (2/3)(x - 4) - 6 = (2/3)x - 8/3 - 6

That doesn't look right.

The general process for finding inverses is as follows:

. . . . .1) Rename "f(x)" as "y".
. . . . .2) Solve for "x=".
. . . . .3) Switch x and y.
. . . . .4) Rename the new "y" as "f^-1(x)".

Try that.

Eliz.