What is the equation of the line tangent to # f(x)=1/(1-x)# at # x=0#?

1 Answer
Jan 30, 2016

#y-1=x#

Explanation:

The slope of a tangent at a point of a function is the derivative of the function at that point, i.e #m=dy/dx|_(x_o)#

So, to find the tangent slope, we differentiate #f(x)=1/(1-x)#
So, #f'(x)=(\frac{d}{dx}(1)(1-x)-\frac{d}{dx}(1-x)(1))/(1-x)^2#

We know that differentiating a constant gives us #0# so the left term on the numerator of the right hand side of the equation is #0#, so #f'(x)=1/(1-x)^2#

Since we're finding the slope at #x=0#, we see that #m=f'(x)=1/(1-0)^2=1/1=\impliesm=1#

Now, the general slope equation is #y-y_o=m(x-x_o)#
We know #m=1#, but we have to find #y_o# and #x_o#. Since the tangent obviously touches the function at where we differentiated the function, we can find them. So, we already know that #x_o=0#, substituting this in the original function, we get #y_o=f(0)=1/(1-0)=1/1=1#

Substituting these values into the general equation, we get the equation for the tangent of the function at #x_o=0#.