Q13... You have the function

(a) From the given graph, you know at least two points that satisfy the function,

and

. Using the first point:

and using the second point,

So, we have two possible solutions:

__Possibility 1:__ which leads to

__Possibility 2:__ which leads to

But these are the same, as

So, both solutions are valid. The solutions on the exam are flawed, however, in that they find

by starting with

without first establishing the value of

.

Part (b) requires you to solve

. The solution will be the same whichever of the previous solutions you choose as they both result in the same

and so have the same equation to solve,

. This can be solved by cases:

__Case 1:__ , in which case

and so

and this solution is valid throughout the restricted domain

.

__Case 2:__ , in which case

as

and so

and this solution is valid throughout the restricted domain

.

So, the complete solution of

is

or

. This can be expressed in interval notation as

.

Part (b) can also be solved graphically by adding the line

to the given diagram, noting that there are intersections at

and

, and then recognising that

when the given graph lies above the line that you have added. The solution

immediately follows.