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
Answer: 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
Answer: 3
