Points of Tangency for Horizontal Tangents in a Function

[mrfunkyg]

Homework Statement

Determine the points at which the graph of the function has a horizontal tangent
f(x)=(x^2)/(x-1)

Homework Equations

The Attempt at a Solution

f'(x)= ((x-1)(2x)-(x^2)(1))/((x-1)^2)
((2x^2)-2x-x^2))/(x-1)^2
f'(x)= ((x^2)-2x))/((x-1)^2)

third step
I set to 0 I get the equation
X^2-2x=0
factor x(x-2)=0
my points of tangency are (0.0) and (2,0)
but the book says it's (0,0) and (2,4)
thanks for the help

 

[fzero]

To find the points, you need to determine (x,f(x)). What is f(2)?

 

[mrfunkyg]

Hello Fzero thank you for the reply.
f'(x)=x^2-2x=0
f'(0)=(0)^2-2(0)=0 (0,0)
f'(2)=(2)^4-2(4)=0 (2,0)

 

[fzero]

f'(x) = 0 is the condition that the tangent be horizontal. The solutions are values of x. To find the (x,y) values of the points you must compute y=f(x), not f'(x).

 

[mrfunkyg]

f'(x) = 0 is the condition that the tangent be horizontal. The solutions are values of x. To find the (x,y) values of the points you must compute y=f(x), not f'(x).

Thank you so much!