C3 functions pt 2

In your g(X), replace the X with f(X) then make it equal to 16 and solve for f(x)

Then consider the two sections of f(x) and see which X values give you what you found

It tells you all the information at the top. If you look at the graph of f you should see that it involves the modulus function.

i know this is the way to do it but i don't know how :/


so the function of that graph is f(x)=|x|-2 ??? but how am i supposed to sub that into g(x) and is doesn't tell me it's a modulus...

|x-2| is not the equation. The gradient is completely different for x<2 and x>2. It's a piecewise function.

Unparseable latex formula:

\displaystyle f(x) = \[ \left\{\begin{array}{ll} x-2 & 2 \leq x \leq 6 \\ -\frac{5}{2} x+ 5 & -2 \leq x< 2 \\\end{array} \right. \]

This is the function $f$ for the domain specified in the question. (I was wrong about the modulus function as I just glanced over the question).
If you're not sure find what value of x needs to be put into g to give 16. Then find what values of x you need to put into f so that f gives the value you need. Hope this makes sense may not have explained it too well.