How do you find the instantaneous rate of change of f(x)=x^2-2/x+4f(x)=x2−2x+4 at x=-1x=−1?

1 Answer
Olivier B.
Jul 6, 2015

At x=-1x=−1, the instantaneous change rate of f(x)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 rarrf(x)=x2−2x+4→ variation rate at x=-1x=−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.

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