# Average rate of change of the function $f(x)=x^3$ over the interval $[x, x+h]$

Apologies that I do not know how to properly format equations etc on this website. My question is fairly self-explanatory as described in the title.

According to the book I'm using, the answer should work out to $3x^2 + 3xh + h^2$. However, after expanding $(x+h)^3$ I reach a denominator of $x^2h + 2x^2h + 2xh^2 + h^2x + h^3$.

I am a bit stumped about how to reach the answer, though I fear I'm making an embarrassingly evident mistake.

I am having a similar problem with $f(x) = {1\over x}$, which works out to $-1\over x(x+h)$ and I am not certain how.

• Welcome to MSE. Please use MathJax. – José Carlos Santos Jul 21 '17 at 16:01
• your denominator should be $h$, and you multiplied incorrectly too – Saketh Malyala Jul 21 '17 at 16:58

You get $$\frac{(x+h)^3 - x^3}{h} = \frac{x^3+3x^2h + 3xh^2 + h^3-x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h}$$

Can you finish it?

You're on the good tracks !

Starting with you denominator :

$$x^2h + 2x^2h + 2xh^2 + h^2x + h^3 = 3x^2h+3xh^2 + h^3$$

now for the rate of change you have to divide that by $h$ so you end up with :

$$3x^2 +3xh+h^2$$

All you had left was factorising some terms and dividing by $(x+h)-x = h$

Now for $f(x)={1\over x}$ :

Let's recall the general equation for rate of change $\Delta f$ :

$$\Delta f = {f(x+h)-f(x)\over h}$$

Now we plug in our $f$ :

$$\Delta f = {{1\over x+h}-{1\over x}\over h}= {{x-(x+h)\over x(x+h)}\over h}$$

$$\Delta f = {-h\over x(x+h)}\cdot {1\over h} = -{1\over x(x+h)}$$