find the equation of the tangent line to the graph of f(x)= 1/x-1 at the point (2,1)?

Answer 1

Assuming that question is:-
f (x) = (x - 1)^(-1)
f `(x) = (-1)(x - 1)^(-2)
f `(x) = (-1) / (x - 1)²
f `(2) = (-1) / 1 = - 1
y - 1 = - 1(x - 2)
y = - x + 2 + 1
y = - x + 3 is equation of required tangent.

Answer 2

If (2,1) is on the curve, the equation is 1/(x-1)

The derivative of that expression is, using the chain rule:
-1/(x-1)^2

This means that the slope at x=2 is:
-1/(2-1)^2 =
-1/1 =
-1

Now you need a line with slope = -1 passing through (2,1):

y = mx + b
y = -1*x + b

When x=2, y=1, we can solve for b:

y = -1*x + b
1 = -2 + b
3 = b

This gives us:

y = -x + 3

Answer 3

Do you know how to derive functions?
If so, here's the formula you'll be using:
y-f(x0) = f'(x) (x-x0), where x0 - the x coordinate of the considered point (2)

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

Your ecuation is now:
y-f(2) = f'(2)(x-2) => y-1 = -x+2 => x+y = 3 => x+y-3 = 0

Note: In your formula, f(x0) can be replaced with y0 if that form suits you most [y-y0 = f'(x0)(x-x0)]

Answer 4

f(x) = a million / (x - a million) f'(x) = - a million / (x - a million)² f'(2) = -a million/(2 - a million)² = -a million = m # slope of tangent y = mx + b plug in m and coordinates a million = -a million*2 + b 3 = b y = - x + 3 or y + x - 3 = 0