What is the linear approximation to f(x) = 2/x at x_0 = 1?

Question

I got
2 - (1/2)x

Answer 1

f(x) = 2/x
f(1) = 2

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

f'(1) = -2

Equation of tangent line: y = -2x + 4

Answer 2

f(x) = 2/x. The derived function is f ' (x) = -2/x^2.
At x=1 the f ' (1) =-2. So the tangent has slope -2.
You want to be at x=1. So the linear approximation around x=1 is the line y = -2x + q.
Since it goes througt (1 , 2) you find 2 = -2 + q or q=4
The linear approximation is y = -2x + 4

Answer 3

No. The derivative is -2/x^2, which is -2 at x_0=1, so you want a line of slope -2. The answer would be
y-2=(-2)(x-1), so
y=-2x+4.

Answer 4

2
2 + h*10^10000
2 - h*10^10000
are all linear approximations of f(1+h).

Answer 5

you favor to charm to the graph of y = one million/x first and use your judgement to charm to a tangent at a aspect close to to 0.254 (perhaps 0.25). The tangent will reduce the x and y axes providing you with 2 factors (0 , 7.9) and (0.5 , 0). Use those 2 factors to get your equation of the tangent (y=mx+c). you need to get something like y = -15.5x+7.9 With this equation, in basic terms replace x=0.254 to get the wanted y value (approx 3.9).