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

1 Answer
Jul 6, 2015

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.