I need help on figuring out qn 3 of this past paper for my upcoming exams

[nocturne jewel]

its on domains and images, which im still unsure about

[azure dew]

We can see if x≥0 the function is undefined so x<0. Domain (-∞,0) image is simple R-{0} BECAUSE we know for 1/x so we can say same for 1/2x.

Second one is just addition multiplication and nesting

@nocturne jewel

[plain marsh]

!nosols

[molten grotto]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[molten grotto]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[nocturne jewel]

So if X<0, the function won't be undefined, and the image is the y value?

[queen star]

$f$ is function written as a fraction involving x in the de nominator

[grand minnow]

Daddy_314

[queen star]

So the first thing to check is for which value(s) of x the denominator might be equal to zero

So we would like to find for which value $x$ we might have
$x - |x| = 0$ so that we exclude these values from the domain of $f$

[grand minnow]

Daddy_314

[queen star]

Do you understand this bit ?

[nocturne jewel]

Okay I think I understand that a bit. So if X>0 , then x- |X| =0 , so that bit won't work to give 1/2x and we need X<0

[queen star]

Yes

By definition of the absolute value,
If x is a positive number, |x| = x

So x-|x| = 0 for all positive numbers x