# Evaluating a Function

• August 27th 2012, 06:54 PM
vaironxxrd
Evaluating a Function
For the function f defined by f(x) = $2x^2-3x$

Evaluate: $\frac{f(x+h)-f(x)}{h}$

I'm solving the problem using a book which actually gives me step-by-step instructions. Thing is I have a confusion with the following distribution, and the book is not showing the distribution step.

$\frac{f(x+h)-f(x)}{h}=\frac{[2(x+h)^2 -3(x+h)]-[2x^2-3x]}{h}$

When I do this step I receive the following:
$\frac{2x^2+2h^2-3x-3h-2x^2-3x}{h}$

What the book gives

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

What am I doing wrong? Everything?
• August 27th 2012, 07:00 PM
topsquark
Re: Evaluating a Function
Quote:

Originally Posted by vaironxxrd
For the function f defined by f(x) = $2x^2-3x$

Evaluate: $\frac{f(x+h)-f(x)}{h}$

I'm solving the problem using a book which actually gives me step-by-step instructions. Thing is I have a confusion with the following distribution, and the book is not showing the distribution step.

$\frac{f(x+h)-f(x)}{h}=\frac{[2(x+h)^2 -3(x+h)]-[2x^2-3x]}{h}$

When I do this step I receive the following:
$\frac{2x^2+2h^2-3x-3h-2x^2-3x}{h}$

What the book gives

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

What am I doing wrong? Everything?

Try these corrections:
1. $(x + h)^2 = x^2 + 2xh + h^2$, not $(x + h)^2 = x^2 + h^2$
2. The sign on the last 3x is positive, not negative

-Dan
• August 27th 2012, 07:09 PM
vaironxxrd
Re: Evaluating a Function
Quote:

Originally Posted by topsquark
Try these corrections:
1. $(x + h)^2 = x^2 + 2xh + h^2$, not $(x + h)^2 = x^2 + h^2$
2. The sign on the last 3x is positive, not negative

-Dan

I saw that before... a very simple mistake. Thank you, for correcting me.