Question

How do you find the instantaneous rate of change of `f(x)=x^2-2/x+4` at `x=-1`?

Answer 1

At `x=-1`, the instantaneous change rate of `f(x)` is null.

Explanation:

When you calculate a function's derivative, you obtain an other function representing the variations of the first function's curve's slope.

A curve's slope is the instantaneous variation rate of the curve's function at a given point.

Therefore, if you are looking for the instantaneous variation rate of a function at a given point, you should calculate this function's derivative at said point.

In your case:

`f(x)=x^2-2/x+4 rarr` variation rate at `x=-1`?

Calculating the derivative:

`f'(x)=(d(x^2))/(dx)-(d(2/x))/(dx)+(d4)/(dx)`

`=2x-(-2/x^2)+0=2x+2/x^2`

Now, you just need to replace `x` in `f'(x)` with its given value, `x=-1`

`f'(-1)=2(-1)+2/(-1)^2=-2+2=0`

The derivative is null, therefore the instantaneous change rate is null and the function doesn't increase or decrease at this specific point.