I got a calculus question, need some help please!?

Question

Find the equation of the tangent line to the curve f(x)=(4/x)-x at the point where x=1

Answer 1

write as y = 4/x -x, and
y' = -4/x^2 -1. at x = 1, the slope of the tangent line is

= -4/1^2 -1
= -4 -1
= -5.

At x =1, y = 4/1 -4
= 4-4
= 0.

So, we have the point (1,0) and a slope of -5. In point-slope form,

(y-0)/(x-1) = -5 or

y = -5(x-1) or
y = -5x + 5. Done.

Answer 2

First find out what f(x) is for the curve. This gives you a coordinate (1, f(x)).
Then take derivitive= -4/x^2-1. Evaluate that at x=1 to get the slope.
You have the slope, you have a coordinate, and the tangent you seek is
F'(x)x=1 = Slope + B, where B= f(x) at x=1.

Answer 3

The slope of a line tangent to the curve is the derivative of that curve. So first to find the derivative:

f(x) = (4x^-1) - x
f'(x) = (-4x^-2) -1

now plug in x =1
-4 - 1 = -5

so now that you have the slope, you just need a point to be able to write an equation. Plug x =1 into the original equation to find that point. 4 -1 = 3

using slope = -5 and the point (1,3)

plug into pt slope formula: (y - 3) = -5(x - 1)
y = -5x + 8

Answer 4

take the derivative of the function to find the slope

derivative = -4/x^2 - 1 = slope

and plug in your x value to get the slope at that point

so slope = -4 - 1 = -5