use the defection of a derivative to determine the derivative\, slope and equation of the tangent line when x=3 y= 2/x

Answer 1

I will interpret this question to mean we need to use the limit definition of a derivative (otherwise, using the power rule for differentiation is quicker and easier):

f^'(3) = lim_x→3 (f(x) - f(3)) / (x - 3)

= lim_x→3 (2/x - 2/3) / (x - 3)

= lim_x→3 (6 - 2x) / 3x) / (x - 3)

= lim_x→3 (-2(x - 3)) / 3x(x - 3)

= lim_x→3 (-2 / 3x) = -2/9 = m_tan |_x=3 through (3 , 2/3). Eqn of tan line: y - 2/3 = -2/9(x - 3)

Answer 2

I assume you meant definition of a derivative.

f(x) = 2/x f(3) = 2/3

f'(x) = lim (f(x+h) - f(x))/h

h→0

= lim ((2/(x+h) - 2/x))/h = lim (2x - 2x -2h)/((x+h)(hx) = lim -2/(x²+hx) = -2/x²

f'(3) = -2/9 = m

Tangent line equation:

y-2/3 = -(2/9)(x-3) = -2x/9 +2/3

y = -2x/9 + 4/3