2 domain/range problems from sample test

[ninjamonke]

Homework Statement

Check the attachment for the problems.
Problem 7 and 8.

Homework Equations

Check the attachment for the problems.
Problem 7 and 8.

The Attempt at a Solution

For problem 7, I did the part a which I got D: (-4, 3] R: [-3.75, 5) and part b (-1, -3). I couldn't understand part c. On part d, I don't know how to find the f(x) equation for f and g. If I had that, I could answer part d. The part e, I know it's g as inverse function. I don't know the domain and range.

For problem 8, I don't understand this whole word problem. If you could help me setup some equation, I could find the area. The same for b and c.----------------------------------------------------------------------
Special thanks to anyone that helps me on this or even does it. This is a sample test of my upcoming test, not a homework. My professor said the real test will be exactly the same except some different numbers. I answered all other questions so far and I just need help with these 2 ones. I'll be using this as my study guide so I'd appreciate if you could answer these w/ explanations.

 

[eumyang]

For problem 7, I did the part a which I got D: (-4, 3] R: [-3.75, 5) and part b (-1, -3). I couldn't understand part c.

Draw the line y = -1 and find the x-coordinates of the points of intersection.

On part d, I don't know how to find the f(x) equation for f and g. If I had that, I could answer part d.

You don't need to do that. The domain of f/g is the intersection of the domains of f and g. However, you'll also need to exclude from this intersection any values of x that make g(x) = 0.

The part e, I know it's g as inverse function. I don't know the domain and range.

If g has an inverse function, then the domain of g(x) = the range of g^-1(x), and vice versa.

For problem 8, I don't understand this whole word problem. If you could help me setup some equation, I could find the area. The same for b and c.

For 8a, let x = the length of the side of the patch that is perpendicular to the building (the vertical side). Note that the total amount of fencing used = the sum of the lengths of the 3 vertical sides (two outer and one dividing) and the horizontal side (parallel to the building). Express the horizontal side in terms of x. Then you can express the area of the patch using A = l*w.

 

[ninjamonke]

Draw the line y = -1 and find the x-coordinates of the points of intersection.

I got that one.

You don't need to do that. The domain of f/g is the intersection of the domains of f and g. However, you'll also need to exclude from this intersection any values of x that make g(x) = 0.

Is the domain (-2, 3]? -2 is where they first intersect. 3 is where f(x) ends but doesn't intersect.

If g has an inverse function, then the domain of g(x) = the range of g^-1(x), and vice versa.

I still don't understand what it is. The domain of g(x) is [-2, 5] and the range is [-2/3, 5]. What do I have to do with theg^-1(x)?

For 8a, let x = the length of the side of the patch that is perpendicular to the building (the vertical side). Note that the total amount of fencing used = the sum of the lengths of the 3 vertical sides (two outer and one dividing) and the horizontal side (parallel to the building). Express the horizontal side in terms of x. Then you can express the area of the patch using A = l*w.

Is that supposed to be y? If so, this is what I did.

A=xy
3x + 2y = 240 (Solve)
y = -3/2x + 120

Now what?

 

[eumyang]

Is the domain (-2, 3]? -2 is where they first intersect. 3 is where f(x) ends but doesn't intersect.

What I mean about intersection is this:
The domain of f(x) is (-4, 3), and the domain of g(x) is [-2, 4.5]. Give me a set that is the intersection of those sets of numbers. Afterwards, you will have to take out any value of x that makes g(x) = 0 (there is one value).

I still don't understand what it is. The domain of g(x) is [-2, 5] and the range is [-2/3, 5]. What do I have to do with theg^-1(x)?

The domain looks more like [-2, 4.5] and the range looks more like [-1.5, 5]. You're being asked only for the domain and range of g^-1(x).
- The domain of g^-1(x) is the same set of numbers as the range of g(x), so the domain of g^-1(x) is [-1.5, 5].
- The range of g^-1(x) is the same set of numbers as the domain of g(x), so...?

Is that supposed to be y? If so, this is what I did.
A=xy
3x + 2y = 240 (Solve)
y = -3/2x + 120
Now what?

The bolded is wrong. The problem states that "no fence needs to be used along the building," so the 2nd line should read
3x + y = 240, so
y = -3x + 240.
Plug this into A = xy, and you'll have an area as a function of the length of the vertical side.