Tutors Answer Your Questions about Functions (FREE)
Question 1000725: We are doing inverses in class and I am stuck on this one problem. Can you help me?
It says,"for the given function f(x), find f^-1(x)
f(x)=x/3 I know I'm supposed to replace f(x) with y and then interchange x and y to get x=y/3 but now i'm stuck. Am I even on the right track? Please help.
Click here to see answer by fractalier(6550)   |
Question 1000732: Trying to find the inverse of this function and I am not having good luck. Could you advise?
For the given function f(x), I am trying to find f^-1(x)
f(x)=-3/4x+6. I start by replacing f(x) with y (y=-3/4x+6) then I interchange the x and y (x=-3/4y+6) from here, I think I am supposed to multiply all terms by 4 (4*x=-3/4x*4+6*4) which would give me 4x=-3y+24. Here is where I need direction. Should I subtract x from both sides? or should I subtract -3y+24 from both sides?
Click here to see answer by stanbon(75887) |
Question 1000824: Which of the following relations is a function?
(8, 1), (-4, 4), (4, 1), (8, 2)
(4, 4), (-4, 6), (4, 3), (-7, 2)
(4, 4), (-4, 2), (8, 1), (-7, 2)
(4, 0), (-4, 3), (8, 1), (-4, 5)
Click here to see answer by MathLover1(20850)  |
Question 1000870: Write a function that models the distance D from a point on the line y = 11 x - 6 to the point (0,0) (as a function of x).
D(x) = √(122x2−132x+36)
The point on the line that is closest to (0, 0) is
( , ) (numerical answers)
I was able to find D(x) but unsure how to find the x and y coordinates? please explain
Thank you
Click here to see answer by josgarithmetic(39618) |
Question 1000952: Is this the right approach to solving limits?
ex:
lim x->∞ (ln(e^x-e^-x))/(x)
lim x->∞ (ln(e^∞-e^-∞))/(∞)
lim x->∞ (ln(e^∞-1/(e^∞))/(∞)
lim x->∞ (ln(e^∞-0))/(∞)
lim x->∞ (ln(e^∞))/(∞)
lim x->∞ (∞(ln(e)))/(∞)
lim x->∞ (∞(1))/(∞)
lim x->∞ (∞)/(∞)
= 1
Thank you
Click here to see answer by ikleyn(52794)  |
Question 1000952: Is this the right approach to solving limits?
ex:
lim x->∞ (ln(e^x-e^-x))/(x)
lim x->∞ (ln(e^∞-e^-∞))/(∞)
lim x->∞ (ln(e^∞-1/(e^∞))/(∞)
lim x->∞ (ln(e^∞-0))/(∞)
lim x->∞ (ln(e^∞))/(∞)
lim x->∞ (∞(ln(e)))/(∞)
lim x->∞ (∞(1))/(∞)
lim x->∞ (∞)/(∞)
= 1
Thank you
Click here to see answer by Alan3354(69443)  |
Question 1000931: Please explain how to find the horizontal and vertical asymptotes of these functions using limits. If I understand correctly, to find the vertical asymptote you set the numerator to zero and solve. And to find the horizontal asymptote you compare the exponents, if they are the same, use the fraction, if the numerator is larger it DNE, if the denominator is larger then it is zero. I understand that decent enough but throwing limits in there is confusing. How do you find vertical and horizontal asymptotes by using limits on these functions?
Thank you
Click here to see answer by ikleyn(52794) |
Question 1001042: Basic question. f(x) equals y because plugging in some value x into the function will give you what the corresponding y-value will be. And the y-value of a LINEAR FUNCTION will always be the same thing as its horizontal asymptote?
I know the function will output differently for a rational function, and then the rules change. but for something like lim x -> ∞ where the function is f(x) = 5+e^(-x^2). By plugging in ∞ into the function this will output the horizontal asymptote because the horizontal asymptote is the same as y-value. In this case y = 5 and that is the horizontal asymptote.
not sure if this is right or not.
Please help!
Click here to see answer by ikleyn(52794) |
Question 1001247: I am having a problem finding the product of this equation. I am using the product function(f*g)(x) f(x)=(x+6), g(x)=x^2-2x+12. After multiplying, I end up with x^3-2x^2+12x+6x^2-12x+72. When I combine like terms, I get x^3-4x^2+72. I was thinking this was my answer but the answer key has the answer as x^2-2x+18. Did I do something wrong? Can you help me?
Click here to see answer by solver91311(24713)  |
Question 1000949: 1. let f(x) = ((2x-1)^2)/(2x^2)
find:
a. lim x->∞
b. lim x->-∞
c. lim x->0
work:
a. lim->∞
((2(∞)-1)^2)/(2(∞)^2) = (4(∞)^2)/(2(∞)^2) = 2. Because we dismiss the -1 and the (∞)^2 cancel eachother out.
b. lim x->-∞
(2(-∞)-1^2)/(2(-∞)^2) = 2. For same reasons.
c. lim x->0
((2(0)-1)^2)/(2(0)) = ((-1)^2)/(0) = DNE because you cannot divide by zero therefore a two-sided limit diverges at zero.
============================
Similar problem:
Let f(x) = ((2x-2)^2)/((x-3)^2)
find:
a. lim x->∞
b. lim x->-∞
c. lim x->3^(+)
d. lim x->3^(-)
work:
a. lim x->∞
((2(∞)-2)^2)/((∞-3)^2)
((2(∞))^2)/((∞)^2))
(4(∞)^2)/(∞)^2 = 4 because (∞)^2's cancel
b. lim x->-∞
similarly, this becomes 4
c. lim ->3^(+)
d. lim ->3^(-)
Bit confused on how to solve these last two. The way I learned it was that 3^(+) approaches 3 from the right, therefore its values is something like 3.00001. And 3^(-) approaches 3 from the left so its value is something like 2.99999. So do I plug and chug these values into the function? or is there another process. One-sided limits confuse me.
Thank you
Click here to see answer by Fombitz(32388)  |
Question 1001656: Let f(x) = x^4 + 2x^3 + 1
I take the derivative and set equal to zero to find its critical pts: x= 0 and x = -3/2. Then found the graph from decreasing(-∞,-3/2]
and increasing [-3/2,0]U[0,∞) "by the way are these brackets correct? is that correct to include the critical pts in the interval?"
Found that the rel extreme is therefore a MIN when x=-3/2
Found its concavity intervals and inflections pts at: x=-1 and x=0.
Concave up: (-∞,1)U(0,∞)
Concave down: (-1,0)
Found the yint at (0,1) but couldn't find the yint. And I guess these cannot actually have H.A. or V.A. that's only for rational functions.
I was able to sort of graph it
here is my graph: http://imgur.com/mFKoUy3
just confused, does the inflection point represent not only where the sign change happens but also a point on which the graph crosses the x-axis? And how does one find the x points from that function to get a complete scope of the graph. I cannot use a calculator.
Thank you!
Click here to see answer by Boreal(15235)  |
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