Question

Find the equation of the tangent line to the graph of the function f(x)=(3x-2)/x+3 at the point (2, 0.8)

Answer 1

f'(x)= 5/(x+3)^2 ===> f'(2)=1/5 ===> y=(1/5)x+0.4

Answer 2

where did you get your 5

Answer 3

i got 11

Answer 4

f(x)= [(3x-2)/x] + 3
f'(x) = [x(3) - (3x - 2)(1)] / x^2 , derivative using quotient rule
f'(2) = (6 - 6 + 2)/4, slope of f at x = 2
= (2)/4
= 1/2

equation of tangent line:

ytan = f'(2)x + b
ytan(2) = .8 = 1(2)/2 + b
b = .8 - 1
= 8/10 - 1
= (8 - 10)/10
= -1/5

Ans. ytan = x/2 - 1/5