Answer:
To find the equation of the tangent line at the point (1,-1), we need to find the derivative of the function y=2/x-3/x^2 first.
y = 2/x - 3/x^2
y' = -2/x^2 + 6/x^3
Now we can plug in x=1 to find the slope of the tangent line at (1,-1).
y'(1) = -2/1^2 + 6/1^3 = 4
So the slope of the tangent line is 4.
To find the equation of the tangent line, we can use the point-slope form:
y - y1 = m(x - x1)
where m is the slope and (x1,y1) is the point (1,-1).
Plugging in the values, we get:
y - (-1) = 4(x - 1)
Simplifying, we get:
y = 4x - 5
Therefore, the equation of the tangent line at (1,-1) is y=4x-5.
Explanation: