Core 3

How do you work out the range of a function?
and when a question says 'for all real values of x' does this mean values bigger than 0?

Range of a function is all of the possible y values. In an y=mx + c eqn then f(x) <_ c
Or you can complete the square if it's a quadratic eqn.

For all real values of x means a rational number. so not sqrt -1 or 36578.4567890 recurring

This is incorrect.

Would it not be more useful to give the correct answer then you absolute salad?

Range of a function is all of the possible y values. In an y=mx + c eqn then f(x) <_ c
Or you can complete the square if it's a quadratic eqn.

For all real values of x means a rational number. so not sqrt -1 or 36578.4567890 recurring

As lizard points out, your range is completely wrong for y = mx + c. If there are no restrictions on the domain of x then this function can 'hit' every real value - just draw a graph of a straight line if you're not convinced!

And "all real values of x" includes irrational numbers, recurring decimals etc. It only excludes complex numbers with non-zero imaginary part.

Can someone please help! I got stuck on this question so in case something like this comes up in the exam tomorrow...

"solve |x|-1=|x^2 - 3|" Now is anyone able to step by step show (and explain why) how to get the 4 solutions wanted (I've checked they are; 2,-2, 0.5(root(17)-1) and -0.5(root(17)-1).

Thank you!

$|x| - 1=|x^2-3|$
$\pm x -1 = \pm(x^2-3)$

Just evaluate each case.

How do I choose which solutions are correct if not told? by evaluating the equations I get 4 extra solutions; 1,-1 and 0.5(1+0r-root(17))

For every solution you get, check each one simply by substituting back into the original equation. If it satisfies the equation, it is a valid solution.

e.g.

$|1| - 1 = 0$
$|1^2 - 3| = |-2| = 2 \neq 0$

So x = 1 is not a solution