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

Jul 6, 2015

At $x = - 1$, the instantaneous change rate of $f \left(x\right)$ 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.

$f \left(x\right) = {x}^{2} - \frac{2}{x} + 4 \rightarrow$ variation rate at $x = - 1$?

Calculating the derivative:

$f ' \left(x\right) = \frac{d \left({x}^{2}\right)}{\mathrm{dx}} - \frac{d \left(\frac{2}{x}\right)}{\mathrm{dx}} + \frac{d 4}{\mathrm{dx}}$

$= 2 x - \left(- \frac{2}{x} ^ 2\right) + 0 = 2 x + \frac{2}{x} ^ 2$

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

$f ' \left(- 1\right) = 2 \left(- 1\right) + \frac{2}{- 1} ^ 2 = - 2 + 2 = 0$

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