SOLUTION: 4. Is f(x) - g(x) = g(x) - f(x) true for all functions? Justify your answer. **Writing one example will not justify your answer for ALL functions. Which operation property can we

Algebra ->  Real-numbers -> SOLUTION: 4. Is f(x) - g(x) = g(x) - f(x) true for all functions? Justify your answer. **Writing one example will not justify your answer for ALL functions. Which operation property can we      Log On


   



Question 1194424: 4. Is f(x) - g(x) = g(x) - f(x) true for all functions? Justify your answer.
**Writing one example will not justify your answer for ALL functions.
Which operation property can we use to prove this is true? **
(2 marks) 5. Is (f + g)(x) = (g + f )(x) true for all functions? Justify your answer.
**Writing one example will not justify your answer for ALL functions.
Which operation property can we use to prove this is true? **

Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Problem 4

One example could be
f(x) = x+5
g(x) = 2x-3

So,
f(x) - g(x) = (x+5) - (2x-3)
f(x) - g(x) = x+5 - 2x+3
f(x) - g(x) = (x-2x) + (5+3)
f(x) - g(x) = -x + 8

On the other hand,
g(x) - f(x) = (2x-3) - (x+5)
g(x) - f(x) = 2x-3 - x-5
g(x) - f(x) = (2x-x) + (-3-5)
g(x) - f(x) = x - 8

This shows that f(x) - g(x) and g(x) - f(x) are not the same.
There are some special cases when f(x) - g(x) = g(x) - f(x) is true for all x; however, the equation is false in general for any functions f,g.

I don't know why your teacher wrote "Writing one example will not justify your answer for ALL functions." when that's exactly all we need. One example is enough to disprove the claim made. The claim is the equation works for ALL functions. But we clearly found two functions where this doesn't work. It seems like your teacher made a typo.

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Problem 5

The reason why f(x)+g(x) is the same as g(x)+f(x) is the same reasoning why x+y = y+x is true.

We can add any two numbers in any order we want
2+3 = 3+2
7+9 = 9+7
etc

Functions are just an extension of this idea.

Here's one example using functions
f(x) = 2x+7
g(x) = 3x-1
f(x)+g(x) = (2x+7)+(3x-1) = 5x+6
g(x)+f(x) = (3x-1)+(2x+7) = 5x+6

The 2x and 3x combine to 5x no matter what order you add them in. The 7 and -1 combine to 6.

As your instructions say, this example does not prove every single case. But it does help illustrate why it works.

The property used here is the commutative property of addition, which was the x+y = y+x mentioned.

Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
.
4. Is f(x) - g(x) = g(x) - f(x) true for all functions? Justify your answer.
**Writing one example will not justify your answer for ALL functions.
Which operation property can we use to prove this is true? **
(2 marks) 5. Is (f + g)(x) = (g + f )(x) true for all functions? Justify your answer.
**Writing one example will not justify your answer for ALL functions.
Which operation property can we use to prove this is true? **
~~~~~~~~~~~~~~~~~~


            Of two problems in the post,  I will consider and solve here
            only one,  namely first one,  which goes under # 4.


The question is "Is identity f(x) - g(x) = g(x) - f(x) true for all functions?".


The answer is "The identity f(x) - g(x) = g(x) - f(x) is not true for all functions."


    +--------------------------------------------------------------------------------+
    |    To prove that the answer is correct and disprove the original statement,    |
    |                                                                                |
    |            it is enough to present one single counter-example                  |
    |                  disproving the original statement.                            |
    +--------------------------------------------------------------------------------+


This disproving counter-example is  f(x) = 1, g(x) = 0,  two constant-value functions,

which produce self-contradictory identity  1 = -1  in real numbers.


The problem is solved,
        the answer is given,
                and the original statement is disproved by presenting one contradicting example.


The instruction,  written in the post between the signs  (**)  and  (**),  that follows the problem,
is non-sensical gibberish,  contradicting standard logic.


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As a post-solution notice and an advise for the future:

            - never pack more than one problem per post.