Question

Question #43d70

Answer 1

-1 or -4, depending on which version of the statement you meant, see below.

Explanation:

The problem as written is a bit unclear as to whether you want the slope of the curve `y=4/(x+1)` or `y=4/x +1`. I will solve both. The first curve shall be `y_1`, the second `y_2`. (`y_2` is the way you wrote the function in the question, but it is possible that you intended to write `y_1`)

For both of these, it is important to note that the slope of the curve will also be the slope of the line tangent to the function at that point, and will thus be calculated by taking the derivative. This becomes easier if we put the function in terms of `ax^n`, which is easily done. Simply recall that `1/x^n = x^(-n)`, and...

`y_1 = 4/(x+1) = 4(x+1)^-1, y_2 = 4/x +1 = 4x^-1 +1`

Then by the power rule...

`dy_1/dx = d/dx 4(x+1)^-1 = -4(x+1)^-2 = -4/(x+1)^2`

`dy_2/dx = d/dx (4x^-1 +1) = -4x^-2 = -4/x^2`

Then calculating for `x=1...`

`dy_1/dx (x=1) = -4/(1+1)^2 = -4/2^2 = -4/4 = -1`

`dy_2/dx (x=1) = -4/(1^2) = -4/1 = -4`

Thus, for `y = 4/(x+1), dy/dx(x=1) = -1`

And for `y=4/x +1, dy/dx(x=1) = -4`