Oops saw this a little too late but when you draw it, do you draw the line going through the x axis and put the open circle on the 2 on the x axis or?
Like I illustrated after that, you take the limit of the x value where the open circle should be, in this case we take the limit of x=-2 because the open circle is at x=-2. If it was at x= k then we take the limit of x=k of the function. When we get a pair of coordinates in this case (-2, 1/4), we plot the open cirlce as just a small open cirlce white inside black outline where that point is, the graph then goes through this open circle.
Wait so why is 2 square root 2 the y co ordinate again?
It can be any coordinate just if it satisfies the function, if I picked x=2, then y would be 2 root 2 times 2. It just helps us in graping the line more accurately since its an inequality graph.
Oh wow thank you so much!
Thank the lord that i'm done with graphing. Now it's time for trig
I need help with this:
Question J please and thank you
Ok for j, we have to first of all extend the domain given, at this moment it is
So we now need to find the domain for 3theta not just theta otherwise we will lose solutions So lets times 3 to every part of our domain
Now lets evaluate it:
We have square rooted and hence on the RHS given us a square root.
Now tan of something is equal to one, then something must be 45 degrees right? (Exact ratios)
If you remember the ASTC method of determining whether tan is positive or negative, then you will know that in first quadrant and 3rd quadrant tan is positive, and in the other ones negative, here we have plus AND minus 1, this means that our theta can be in ANY quadrant, so we account for all quadrants because it is plus minus.
So now we follow the ASTC rule and the 180-theta, 180+theta etc. etc. rules
Now divide 3 to everything
And the final solutions are all those values above (all in degree measure, it would be a pain to write \circ for each one)
