2.8, #2, Given Problem.

Use the graph to estimate the value of each deriva- tive. Then sketch the graph of f '.	2.8, #2, Given Problem.
	We approximate a tangent line to the curve at x = 0. We then estimate the slope of that line. In the estimate on the left, the slope of the black tangent line appears to be about -3. Therefore, f '(0) is approximately -3.
	We do the same thing at x = 1, 2, 3, 4, 5. Below are my estimates for the slopes of these lines: The slopes in this chart, then become our estimates for the derivative values at each value for x. I added the 2.5 value because the slope seemed slightly larger than at x = 2 and x = 3.
	If we plot each of those slopes as points and then approximate the curve, we get the blue colored result at the left.

2.8, #22, Given Problem.

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.	2.8, #22, Given Problem. The domain of this function has to exclude negative numbers because of the root. Therefore, we could state our domain as: . In interval notation this is: .
	This is how we have learned to calculate a derivative.
	We calculate the derivative using the defined function from the problem.
	Simplify the numerator.
	We separate the fraction into two separate fractions.
	We simplify the second fraction, multiplying both top and bottom by the conjugate of the numerator.
	This gives us the difference of two squares in the numerator.
	Simplify the numerator.
	Cancel the h's and now we can find the limit. The domain of this new derivative function must exclude zero because of the variable in the denominator, and must exclude negatives because of the variable under the root. Therefore, the domain of the derivative function is: . In interval notation this is: .

2.8, #32, Given Problem.

	2.8, #32, Given Problem.
g is discontinuous at x = -2, 0, and 5.	We can see the discontinuities at these 3 points by looking at the graph. At x = -2 and x = 5, we clearly have gaps in the graph.
g is not differentiable at x = -2, 0, 2, and 5.	Any time a graph is discontinuous, it cannot be differentiated. This takes care of x = -2, 0, and 5. Because there is a cusp at x = 2, where no slope can be determined, this too is a place where the function is not differentiable.

	Assignment
1-43, Odds
Lesson 2.8, Pages 167-170