Posted by Jasmin on Saturday, June 13, 2009 at 1:41am.
You need to add parentheses to define the numerator and denominator.
Do you mean g(x) = (10-x)/(2+x), or
10 - (x/2) + x ?
Other interpretations are also possible. They are not the same function
10-x is on the top and 2+x is on the bottom then it asks for
(a) Find g (–1).
(b) State the domain of the function .
(c) Find g (t + 1) and simplify as much as possible.
Does this ans your ? I am really confused and need help. Thanks!
I don't think drwls is still here, so I will reply instead.
so we have
(gx)= (10-x)/(2+x)
then g(-1) = (10-(-1))/(2+(-1))
= 11/1 = 11
(notice is simply replaced the x with -1 wherever it showed up)
the domain is simply your choice of numbers for x which will result in a real number.
so in (10-x)/(2+x) we can sub in any value of x except x=-2 which would result in the denominator equal to zero.
Of course a division by zero is undefined, thus no real number answer.
for g(t+1) we replace any x we see with t+1
so g(t+1) = (10 - (t+1))/(2+(t+1))
= (9-t)/(3+t)
Why do we use t as a variable? What should I be looking for as I figure this out?
Is t only used in part c?
your question was
<< (c) Find g (t + 1) and simplify as much as possible. >>
so I replaced x with tt+1 and simplified
e.g. g(☻) = (10-☻)/(2+☻)
So the domain is any set of real numbers? What would the domain part look like?
for "a" g(-1)= 11
and g (-1) = (10-(-1))/ (2+ (-1))
= 11/1 = 11
it depends on your equation.
for a function of the form
f(x) = 'any algebraic expression containing some x's'
the "domain" is the set of all the x's you can choose from to obtain a real number as an answer.
the "range" is the set of all those real numbers you get, or all those f(x) results.
The most common problems with domain arise when ..
1. there is a denominator which turns to zero for some value or values of x
e.g. in yours there is a denominator of (2+x)
Doesn't 2+x become zero when x=-2
That is why your domain is the set of all real numbers except x=-2
2. if your equation contains a square root of some expression, which becomes negative for some value or values of x
Since we cannot take the square root of a negative number, all those guilty x's have to be exclude.
e.g. f(x) = √(x+5)
all x < -5 would cause a problem, so the domain is :set of all real numbers, except x < -5
So for mine the answer would be
(-infinity,-2) U (-2,infinity)?
not familiar with your notation for domain
in the good ol' days we would have written this domain as
domain = {x │ x (the 'is an element' symbol) R, x ≠ -2}
So is my answer correct?
(-infinity,-2) U (-2,infinity)?
Did you not read my reply??
your answer of (-infinity,-2) U (-2,infinity) contains the UNION symbol
which means any element in the first set OR any element in the second set.
I would conclude you want -2 to be included.
But that would be wrong!
the -2 has to be excluded.
So what if i take the U symbol away and leave -infinity, -2 and -2 infinity. would this be correct?