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Hello,
I can't solve these two exercises.
C Are f and g inverses of each other? Support your response algebraically.
8 f(x) = -(⅕)x + (⅗), g(x) = -5x + 3
9 f(x) = x + ½, g(x) = x - ½
Thank you so much,
Jade
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Have you tried reversing x and y in one and then rewriting it as an f(x) ?
That's the usual way.
World Peace Thru Frisbee
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hi Jade.
If you make a graph of a function you'll get a certain graph. If you graph the inverse function the graph you get is a reflection of the first in the line y=x. The picture below shows this happening.
Why? Well when you input an x value into the first function you get a certain y value. If you treat that as an x input in the inverse function the y output that you get must be the original x you started with, because that's what an inverse function does.
I'll illustrate with some numbers:
Start with x = 10. y = -10/5 + 3/5 = -7/5
Now use that as an x input in the inverse function:
x = -7/5 y = -5 times -7/5 + 3 = +7 + 3 = 10 which brings me back to the number I started with.
One way to show a second function is the inverse of a first function would be to try out every possible x as input and do what I've done above. But you couldn't try every number this way so we need something better.
The graph shows a way to check if the second function is the inverse but graphs aren't usually acceptable as proof since they can never be 100% accurate.
The method suggested by Phrzby Phil is one way to make a proof and it relies on the y=x symmetry. By swopping x and y in a function you are reflecting the graph in the line y=x. So you are creating the inverse function.
Function: y = -x/5 + 3/5
Swop x and y.
x = -y/5 + 3/5
Now make y the subject.
Times all by 5 to remove the fractions
5x = - y + 3
Add y to each side.
5x + y = 3
Subtract 5x from each side.
y = -5x + 3
Another way to prove it would be to use composite functions. If you apply the first function, and then apply the second to the result the simplified result gf(x) should be just x.
f(x) = -x/5 + 3/5
g(-x/5 + 3/5) = -5 times (-x/5 + 3/5) + 3 = (x - 3) + 3 = x
If g(x) is the inverse of f(x), then f(x) is also the inverse of g(x). I suggest you try the following for practice.
Start with g(x)
Swop x and y; make the new y the subject and simplify. You should end up with f(x)
Also work out fg(x) using composite functions. Here again you should end up with just x.
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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Hello,
Thank you for your help! This is what the teacher is telling me about how to do these exercises.
Step 1 Find f(g(x)).
Step 2 Find g(f(x)).
Step 3 If both Step 1 and Step 3 have x as the result, then they are inverses.
The problem is that I can't solve the first two steps for exercises 8 and 9.
Jade
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hi Jade
Another way to prove it would be to use composite functions.
And I have shown all the algebraic steps for Q8 g(f(x)). I'll show one more:
Q9. f(x) = x + 1/2
g(x+ 1/2) = x + 1/2 - 1/2 = x + 0 = x
Thus applying f and then g brings us back to x, which is what is meant by an inverse.
Hopefully you can do f(g(x)) for each yourself.
If p(x) is the inverse of q(x) then q(x) is automatically the inverse of p(x), so it is not strictly necessary to do both fg and gf. Either one is sufficient to prove an inverse. You couldn't have an pair of functions where gf(x) = x but not fg(x) = x. But, if they are asking for both then you'd better do both. As I've done fg in both questions, doing gf yourself will be a good test that you can do this work.
Worth studying: https://www.mathsisfun.com/sets/function-inverse.html
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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Hello,
Thank you so much for your detailed explanation!
This is what I came up with f(g(x)) for exercise 8.
f(g(x))= f (-5x + 3) = -1/5 (-5x + 3) + 3/5 =
(After that I am stuck and don't know how to get x as the result.)
This is the result for f(g(x)) for exercise 9.
f(g(x))= f (x -1/2) = x - 1/2 + 1/2 = x - 0 = x
(IS THIS CORRECT?)
Jade
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Sounds to me that you are saying you don't know the distributive rule: a(b + c) = ab + ac.
That's pretty fundamental. You should know this before doing inverse functions.
If you DO know it, then use it here: -1/5 (-5x + 3)
and then simplify further.
Last edited by Phrzby Phil (2022-08-14 09:44:40)
World Peace Thru Frisbee
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hi Jade,
Phrzby Phil is right. You need to practise some algebra:
https://www.mathsisfun.com/algebra/expanding.html
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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Hello,
I have one last exercise I need to check if it is correct.
Dwight will be celebrating his 7th birthday with a swimming party that his parents are planning for him. The facility rents for $500 plus $25 per guest including lunch and drinks. How many guests can they invite if their budget is $2500?
2500 - 500 = 2000
2000/25 = 80 guests
Thank you for your help,
Jade
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Hello,
Does this look correct considering the feedback in the parenthesis?
10. Dwight will be celebrating his 7th birthday with a swimming party that his parents are planning for him. The facility rents for $500 plus $25 per guest including lunch and drinks. How many guests can they invite if their budget is $2500?
(Show the equation substituting into y and then how to solve the equation for x)
y = mx + b
y = 25x + 500
2500 = 25x + 500
25x = 2000
x = 80 guests
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Hi Jade,
That looks like what is expected.
If I were answering this I would add (after y = mx + b)
where x is the number of guests,
b the fixed charge,
m the charge per guest
and y the budget available.
That way it shows what each leter stands for, which helps someone to follow what you are doing.
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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