When #x=3# we have;
# f(3)=((3-2)(3-1))/(e^3-1) = 3/(e^3-1) #
# f(x) = ((x-2)(x-1))/(e^x-1) #
# :. f(x) = ((x-2)(x-1))/(e^x-1) #
# :. f(x) = (x^2-3x+2)/(e^x-1) #
We now use the quotient rule: # d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #
And so;
# f'(x) = ((e^x-1)(2x-3) - (x^2-3x+2)(e^x))/(e^x-1)^2#
We don't need a more simplified expression, we just need to find #dy/dx# when #x=3#;
# :. f'(3)=((e^3-1)(6-3)-(9-9+2)(e^3))/(e^3-1)^2 #
# :. f'(3)=(3(e^3-1)-2e^3)/(e^3-1)^2 #
# :. f'(3)=(3e^3-3-2e^3)/(e^3-1)^2 #
# :. f'(3)=(e^3-3)/(e^3-1)^2 #
So the tangent has gradient # m=(e^3-3)/(e^3-1)^2 # and passes through #(3, 3/(e^3-1) )#, so using #y-y_1=m(x-x_1)# the required equation is given by:
# y-3/(e^3-1)=(e^3-3)/(e^3-1)^2 (x-3) #
# :. y=(e^3-3)/(e^3-1)^2 x -3((e^3-3))/(e^3-1)^2 + 3/(e^3-1)#
# :. y=(e^3-3)/(e^3-1)^2 x + (-3(e^3-3) + 3(e^3-1)) / (e^3-1)^2 #
# :. y=(e^3-3)/(e^3-1)^2 x + (-3e^3+9 + 3e^3-3) / (e^3-1)^2 #
# :. y=(e^3-3)/(e^3-1)^2 x + 6 / (e^3-1)^2 #
