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Hello,
Can I get feedback on if these look correct?
1 Given f = { (-1, 8), (3, -1), (-2, 0) } and g = { (-5, 8), (3, -1), (-1, 0) } . Find (f o g)(3). Show steps,
( f o g)(3) = f ( g(3) ) = f(-1) = 3 (IS THIS CORRECT?)
2 Given f = { (-1, 8), (3, -1), (-2, 0) } and g = { (-5, 8), (3, -1), (-1, 0) } . Find (g o f)(3). Show steps,
( g o f)(3) = g ( f(3) ) = g(-1) = 3 (IS THIS CORRECT?)
5 f(x) = 15x -12 and g(x) = -15x^2 + 14x - 10, find (f o g)(-1).
( f o g)(-1) = f ( g(-1) )
g(-1) = -15^2 + 14(-1) - 10 = -249
f(-249) = 15(-249) - 12 = -3747
(f o g)(-1) = -3747 (FOR THIS ONE I WAS TOLD THAT -15 IS NOT SQUARED, BUT IN THE G FUNCTION IT IS SQUARED. WHAT AM I SUPPOSED TO CHANGE?)
Is it supposed to be done like this:
(f o g)(-1) = f( g(x))
f( -15^2 + 14x - 10)
15x ( -15^2 + 14x - 10) - 12
-3375x + 210x^2 -150x - 12
= 210^2 -3525x - 12
6 f(x) = -13x2 -13x + 14 and g(x) = -13x - 11, find (g(g(-3)).
(g o g)(-3) = g ( g(-3))
= g ( (-13x - 11) )
= -13 (-13x - 11) -11
= -169x + 143 - 11
= 169x + 132 (THE FEEDBACK WAS THAT FIRST, g(-3)) PUT -3 IN THE G FUNCTION AND THEN SIMPLIFY)
Is it supposed to go like this:
(g o g)(-3) = g ( g(-3))
g(-3) = -13(-3) -11 = 28
(g o g)(-3) = 28
7 f(x) = 2x + 3 and g(x) = –x^2 + 5, find (g o g)(x).
(g o g)(x) = g (g(x))
= g (-x^2 + 5)
= √(-x^2 + 5)
= -x + 5 Domain: {x:x Є R} (I WAS TOLD THAT THE RADICAL SHOULDN'T BE THERE. HOW DO I SOLVE THIS?)
8 f(x) = x + 3 and g(x) = √x, find g(f(x)).
(g o f)(x) = g ( f(x))
= g (x + 3)
= √(x + 3) Domain: {x: x ≤ 3} (IS THIS CORRECT?)
9 f(x) = x - 1 and g(x) = 3/(x-1), find f(g(x)).
(f o g)(x) = f (g(x))
= f [ 3/ (x - 1) ] Domain: {x: x ≥ 1} (IS THIS CORRECT?)
FOR THE FOLLOWING EXERCISE I HAVE SOLVED PARTS A AND B, BUT AM TOLD THAT THE ANSWER FOR PART C IS NOT ENOUGH.
10 You go to a local mechanic to get your tires changed. The tires cost x dollars. There is a 6% sales tax, but you get a 10% discount.
A Write a function t(x) for the total purchase amount after taxes but before discounts and fees.
Function with no discount and fees:
tax: 6%
tire cost: x
t(x) = x + 6% of x
t(x) = x + 0.06x
t(x) = 1.06x
B Write a function d(x) for the total after discounts on purchase amount x but before taxes and fees.
Function with no taxes and fees:
When discount is given after tax, then 10% is deducted from the original cost of the tire:
d(x) = x - 0.10x
C Does it make a difference in the total price whether the mechanic adds the tax first d(t(x)) or takes the discount first t(d(x))? Do not replace x with a numerical value. Show the work to support your answer while keeping x in your work.
It does make a difference in the total price if discount is applied first. Out of the whole amount, you would pay 90% of what it costs. When taxes are added to the discounted price, then the total price would be way less than if the taxes were added first and then the discount.
Thank you so much,
Jade
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Hi Jade,
Let me start with a bit of function theory. In these examples you start with a number, the input, the function does something to produce another number, the output.
Eg.for the square function,an input of 30, gives an output of 900.
For the Sine function an input of 30 gives 0.5
To describe how it does this you might have a formula, eg f(x) = 2x + 3
Or you could have a graph where the x values are for the inputs and the y values are the output,
Or, as here by giving the x and y coordinates where the first is the input and the second the output.
Mathematicians are not consistent about which to put first. For sine the function comes before the input eg. Sine(30) but we don't say square(30) but rather the input is followed by the function as x squared. Sorry it's confusing but you just have to get used to it as it's been like that for a long time.
In composite function work the function always comes before the input which is why fg(3) means start with 3, use the function g to get an output, then use that as input into f.
Now for the questions:
(1). g(3) = -1 is correct for the first stage,but that output of -1 becomes an input for f so we need to look for a coordinate starting (-1,?) so your final answer is not correct.
(2). In the same way -1 is an input for g so look for (-1,?)
(5). You have got in a muddle with -15 x^2
This means square x then multiply it by -15.
You have treated it as - (15x)^2 You shouldn't besquaring the 15 at all.
(6) To do gg(3) you have to start with 3, apply g to it and get an answer. Then start with that value and apply g to it for the final result.
(7) Same thing here but with algebra rather than numbers.
So g(x) = - x^2 + 5 then gg(x) = -(-x^2+5)^2 + 5 You could expand that bracket and simplify.
(8) is correct.
(9) Tricky for me to show this as I cannot do fancy math notation at the moment. I'll comeback to this one later in the week.
(10). Parts A and B look ok to me.
Although it says don't use a number I suggest you do exactly that. You should find the results are the same.
Why? D(x) simplifies to 0.9x. T(x) = 1.06x
Doing one,followed by the other is
Either 0.9 times 1.06 times x
Or 1.06 times 0.9 times x
Both give the same answer.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi Jade,
I'm back to using my laptop now so I can make better looking posts.
If a function has one input I think of it like this:
You 'plug' a number into the box and another number 'pops' out the other end of the box.
For question 9 the function g is more complex so it might help if you think of it like this:
First start with x. Then take away 1. Function so far
Then we have to put that expression on the bottom of a fraction. This is known as taking the reciprocal. Function is now
The final step is to make the result three times bigger by multiplying by 3. Final function
We are asked for fg(x). The above is g(x) so now take one from this:
You could leave this as the answer, but it can be simplified by putting all over the same denominator:
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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