1. if f(x) = 3a|2x – 4| – ax, where a is some constant not equal to zero, find f ′(2). dne 1 0 not enough information 2. find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2. y = –2x + 2 y = –x + 12 y = x + 2 y = –2x – 12 3. use your graphing calculator to find the value of f ′(1) for f(x) = x2 + ln(x). 1 3 2 0

1. If f(x) = 3a|2x – 4| – ax, where a is some constant not equal to zero, find f ′(2).

At x = 2 the function thre is a pointed vertex, because the function 2x - 4 changes of sign, so the derivative does not exist, and the answer is DNE.

2. Find the equation of the line tangent to y = –x^2 + 3x + 8 at x = 2.

1) find the slope, m,  of the equation

m = y' = -2x + 3, evaluated at x = 2

=> m = -2(2) + 3 = -4 + 3 = - 1

2) find the y-coordinate of the point at x = 2

y = -x^2 + 3x + 8 = - (2)^2 + 3(2) + 8 = -4 + 6 + 8 = 10

=> point (2,10)

3) use the point-slope equation:

y - 10 = - 1 ( x - 2)

y - 10 = - x + 2

y = -x + 2 + 10

y = -x  + 12

y = –x + 12

3. Use your graphing calculator to find the value of f ′(1) for f(x) = x^2 + Ln(x).

you can find the derivative: f'(x) = 2x + 1/x

=> f'(1) = 2(1) + 1/1 = 2 + 1 = 3

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