Question

How do you find the intervals of increasing and decreasing using the first derivative given `y=(x+4)/x^2`?

Answer 1

`y(x)` is strictly decreasing in `(-oo,-8)` and reaches a local minimum in `x=8`, then it increases in `(-8,0)`, and it is decreasing again in `(0,+oo)`

Explanation:

Evaluate the first derivative of the function:

`dy/dx = d/dx((x+4)/x^2 ) = (x^2-2x(x+4))/x^4 = -(x+8)/x^3`

and solve the inequality:

`dy/dx < 0`

The inequality is satisfied when numerator and denominator are both positive or both negative, so:

`dy/dx < 0` for `x in (-oo,-8) uu (0,+oo)`

and:

`dy/dx > 0` for `x in (-8,0)`

while the only stationary point is for `x=-8`.

So `y(x)` is strictly decreasing in `(-oo,-8)` and reaches a local minimum in `x=8`, then it increases in `(-8,0)`, and it is decreasing again in `(0,+oo)`

graph{(x+4)/x^2 [-1.1181, -6.118, -0.86, 1.64]}

graph{(x+4)/x^2 [-9, -7, -0.08, -0.04]}