Method 1: First off, you can use the zero function on your calculator by pressing 2nd calc, zero. If the calculator comes up with a zero, then there’s a root there. But in this case, an error comes up saying ‘no sign change’. If this happens, then either there is no root, or the root touches the x-axis but does not cross it. This can be seen, for example, with the graph f(x)=x2.
Method 2: So we still need to figure out whether there is a root or not. We can use the zoom-in function on the calculator, closing in on the root until we can clearly see whether it touches the x-axis or not. (I don't know if you can really tell, but I promise you that it's not touching the x-axis in this diagram. Cross my heart.)Method 3: Finally, we can find the minimum. Press 2nd, calc, minimum. The procedure from there is the same as if you were calculating a zero. (First pick a point to the left, then to the right). If the minimum value is greater than zero, then there is definitely no root. I personally think this is the most efficient way to complete the problem.
b) This function was very sneaky… it appears at first to have three roots. However, by looking at the equation we can tell that there are actually 4 roots. The highest exponent is even, so we know that both outer arms of the graph must go in the same direction. In this case, there was a negative sign in front of the function, so we know they point downward. Therefore there must be another root somewhere off to the right. We can find this root two ways.
Method 1: Expand the size of the window until you see the root. Then it is just a case of using the zero function on your calculator.
Method 2: Go to the table by entering 2nd, table and find the zero. Well, in this case it’s a nice prettyful zero at 50, but of course we will not always be that lucky. Just look for where the y-values go from positive to negative, and you will have a good idea where to look for the zero. Then it’s just a case of using method 1 to find your root.
I'm not going to go into the rest of the quiz, since I don't have the questions. After the quiz we started to talk about inverses, which is going to lead into a lesson on logs tomorrow (I think). First of all we discussed that exponents are the inverse of logs. Mr. K said that logs of base 10 and e are the most frequently used, and that after high school they are pretty much the only thing we will see. So here are some of the ideas we went over:
1) Find the inverse of the function:
Method 2) Switch x with y in the equation and then isolate y. You'll get the same answer as stated above. Note that this method won’t work on all functions. For example, f(x)= (2x-1)/(x+3) will not work. The reason for this is that 2 things are happening to x at once.
2) We have 2 functions: f(x)=2x+3 and g(x)= ½(x-3). What is f(g(x))?
f(g(x))= 2(1/2(x-3)) +3
In this case what you put in is what you get out. In other words, these two functions are inverses.
And that’s all I have guys. Stay tuned for tomorrow’s thrilling continuation! And the lucky winner of the ‘next scribe’ competition is…. Jann!