Question

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

Answer 1

`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`.