I thought the most interesting problem in that packet we were working on for claims was number 46. I’m not exactly sure what it has to do with limits, but piecewise functions were always kinda fun. This problem dealt with transforming the signum funtion. This piecewise function only has 3 outputs. If x is less than zero, the output is negative one. If x is zero, the output is zero. If x is greater than zero, the output is positive one. This function is useful in computing for testing a value ov x to see what sign it has. The graph of the signum function looks like this:

The question first asked about the limit and continutity of the function if r(x) = |sgnx|. Because absolute value makes all imputs positive, it means all values of the function except for zero would be one. So yes, the funtion does have a limit because from both sides as x approaches 1, both outputs are at 1. We don’t care about the value at zero. It does have a function value at zero, which is zero. It is not continuous at zero because the defined point does not line up with the rest of the function.

Next, the question asked to sketch a graph of a transformed piecewise function. It looked like this:

The 3 causes a vertical dilation while the subtraction of two from the imputs causes  a horizontal transformation to the right by 2 units.

Then the question asked to sketch a graph of g(x) = x^2 – sgnx. The graph looks like this:

For part d, the problem asked you to prove that the function a(x) = |x|/x is equal to x for all x except zero. To show this, I did three calculations:

a(1) = |1|/1 = 1 and sgn(1) = 1 because according to the function, sgn of any x greater that one is 1

a(-1) = |-1|/-1 = -1 and sgn(-1) = -1 because according to the function, sgn of any x less than one is -1

a(0) = |0|/0 which is undefined and sgn(0) = 0 so just as the problem said, this doesn’t match. The above calculation will also work for every number because any number divided by itself will be one and the absolute value divided by itself will always be negative one.

Finally, the problem asked you to sketch the graph of f(x) = cosx + sgnx, which looked like this:

We know what cosine looks like without any transformations. When the signum function is added to it, 1 is added to all of the points with x values greater than zero. -1 is added to all of the points with x values less than zero. This causes a vertical translation to the function. It looks like the function has changed at zero because the signum function is piecewise. So on the right, it moves 1 up and on the left it moves 1 down. There is still a point at (0,1), which is the same zero for the cosine function.

That’s all there was to the problem. It has incorporated knowledge we’ve aquired throughout this entire year. This was one of the exploration problems so though it didn’t have much to do with limits, this still taught us something. For more practice on these types of problems, see the packet this came out of and other related worksheets.