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**ziabing****Member**- Registered: 2021-09-18
- Posts: 5

Hello!

I'm struggling with the following question :

True or false?

If f,g : R → R are defined as f(x) = |x−1| et g(x) = |x+1| then the composite function g ◦ f verifies, for x ∈ R :

(g ◦ f)(x) = 2-x if x ≤ 1,

(g ◦ f)(x) = x if x > 1,

I'm pretty sure that (g ◦ f)= |x-1|+1. However, I don't know how to continue. I've graphed all three functions and the statement seems true to me, but I'm not very confident.

Can someone confirm if the statement is true or false? If it's true, how can I prove it? If it's false, how can I disprove it?

Thanks in advance!

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi ziabing

Welcome to the forum.

Here's how I tackled this.

I drew a pair of function boxes,; the first showing f; the second g.

Then I tried values of x, starting with x = 1, then 2 then 3 etc.

1 .........f .........0 ......g ......1

2 .........f .........1 ......g ......2

3 .........f .........2 ......g ......3

This is certainly gf (x) = x for x ≥ 1

So far so good.

0 ..........f .........1 ......g ......2

-1 .........f .........2 ......g ......3

-2 .........f .........3 ......g ......4

-3 .........f .........4 ......g ......5

That looks good too.

How do you 'prove it'.

For x ≥ 1, f(x) = x-1

So gf(x) = g(x-1) = x-1 + 1 = |x| = x as x is positive in this range.

For x < 1, f(x) = 1 - x

So gf(x) = g(1-x) = |1-x + 1| = |2-x| As x < 1, 2-x is always positive so the absolute lines are unnecessary, hence gf(x) = 2-x

Hope that helps,

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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**ziabing****Member**- Registered: 2021-09-18
- Posts: 5

Understood. Thank you so much!

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