Are you in Calculus?
Did you mean ...
f(x) = (2x-3)/(x-2) ?
then f'(x) = ((x-2)(2) - (2x-3)(1) )/(x-2)^2
= (2x-4 - 2x+3)/x-2)^2
= -1/(x-2)^2
It is easy to see that this expression will be negative for all values of x, x ≠ 2
What have you learned about a function when its derivative is negative?
2. Again, I am sure you meant
f(x) = 10/(x+3)
Get a "feel" for numbers, as x --> +∞ , the denominator becomes +∞,
so 10/huge ---> 0 but would be slightly above the x-axis
as x --> -∞ , the denominator becomes
10/(-huge) --->0 but below the x-axis, (negative)
Can you draw your conclusion from that?
Can someone help me with these.
1.Name any intervals on which f(x)=2x-3/x-2 is increasing or decreasing.
2. Find any horizontal asymptotes of f(x)=10/x+3.
4 answers
I am in pre-calculus, and my teacher barely explains the content...
As for question 1, there is no brackets in 2x-3 or x-2, I am not sure if it will make a difference to it
As for question 1, there is no brackets in 2x-3 or x-2, I am not sure if it will make a difference to it
I have to rely on youtube and jiksha >.<
I am sure that in your book the 2x-3 is the numerator and the x-2 is the denominator.
In that case you HAVE to use brackets to force that order of operation.
e.g. if x = 6, the way I read it, the value would be
(2(6) -3)/(6-2) = 9/4
subbing x=3 into your typed expression gives me
2(6) - 3/6 - 2
= 12 - 1/2 - 2
= 9 1/2
so, of course it matters.
In that case you HAVE to use brackets to force that order of operation.
e.g. if x = 6, the way I read it, the value would be
(2(6) -3)/(6-2) = 9/4
subbing x=3 into your typed expression gives me
2(6) - 3/6 - 2
= 12 - 1/2 - 2
= 9 1/2
so, of course it matters.
1 answer