2.7, #8, Given Problem.

	2.7, #8, Given Problem.
	We use the definition of a derivative and the given function to begin calculating the derivative.
	We cube the (x+h) quantity.
	Gather like terms i the numerator and factor out the h.
	After cancelling the h, we find that we can now calculate the limit. We have found the derivative of function g.
so,	Next, we need to determine the value of the derivative at x = 0. (f'(0))
tangent line equation:	We now know the slope of the tangent line at x = 0. Using the slope-intercept form of a line, we substitute in our given point (0,1) and find that the y-intercept is 1. Now using the slope (0) and the y-intercept (1), we have the equation of the tangent line to function g at the point (0,1).

2.7, #24, Given Problem.

This limit represents the derivative of some function f and some number a. State f and a.	2.7, #24, Given Problem.
The form of the derivative that we need here is:	The hint which tells us to used this form of a derivative is the t - 1 in the denominator.
and	a is rather obvious as 1 in the expression. The only subtracted quantity in the numerator is 2. If 2 is f(1) then we have it. And it is... So .

2.7, # 36, Given Problem.
 

For the above function, determine whether f '(0) exists.

2.7, # 36, Given Problem.

We first examine the functon,  by looking at its graph. It appears as if the left-hand limit agrees with the right-hand limit. We also know that at zero, the function would be undefined. However because this function is defined as a piece-wise function, the function is defined at x = 0. Because the function is defined at x = 0, the function is continuous as defined, and the left-hand and right-hand limits agree, f '(0) does exist.

	Assignment
1-31, Odds
Lesson 2.7, Pages 155-156