Need help! For f(x)=|x-2| + |4-2x|, how do we derive it to f(x) = k|x-2|?

Question

What is f(x) when x > 3 and when x < 1?

Find value of k.

Answer 1

k = 3

Here's why:

Definition: |a| = a (when a >=0) and |a| = - a (when a < 0).

When x = 2, x - 2 = 0 and 4 - 2x = 0 so we have 2 cases:

1.) x < 2
Then x - 2 < 0 and, by definition, |x-2| = - (x - 2) = 2 - x
and 4 - 2x > 0 and, by definition, |4 - 2x| = 4 - 2x
so f(x) = 2 - x + 4 - 2x = 6 - 3x = -3(x - 2) {equation1}

2.) x>2 or x=2
Then x - 2 > = 0 and, by definition, |x-2| = x - 2
and 4 - 2x < = 0 and, by definition, |4 - 2x| = - (4 - 2x) = 2x - 4
so f(x) = x - 2 + 2x - 4 = 3x - 6 = 3(x - 2) {equation2}

From {equation1} and {equation2} we can see that
f(x) = 3 · | x - 2|

___

f(x) = 3(x - 2) when x > 3 (from {equation2})
f(x) = - 3(x - 2) when x < 1 (from {equation1})

___

Hope this helps!

Answer 2

first recall that |x - a| = |a - x| since absolute value is the distance between the two values

so rewrite it as f(x) = |x - 2| + |2x - 4|

factor out a 2 |x - 2| + 2|x - 2| = 3|x - 2| so k = 3

i'm not sure what you mean when x > 3 and x < 1
f(x) is always positive if that is what you're asking

Answer 3

|4-2x| = |2x-4| = 2*|x-2|

So f(x) = 3*|x-2|

When x >3, f(x) = 3*(x-2)

When x<1, then the term inside the modulus becomes -ve
So |x-2| = 2-x
f(x) = 3*(2-x)

Answer 4

f(x)=|x-2| + 2|2-x|
if x>3 then
f(x)=x-2 - 2(2-x)=x-2-4+2x=3x-6

if x < 1 then
f(x)=-(x-2) + 2(2-x)
=-x+2+4-2x=-3x+6