What is the equation of the line tangent to f(x)=1/(1-x)f(x)=11x at x=0x=0?

1 Answer
Jan 30, 2016

y-1=xy1=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)m=dydxxo

So, to find the tangent slope, we differentiate f(x)=1/(1-x)f(x)=11x
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.