Simplify the difference quotient of f(x)=-8 + 2 x^2 ?

So, here's the function: f(x) = 2x² - 8

Here is the formula definition of the difference quotient:

f(x+∆x) - f(x)
---------------------
∆x

For f(x+∆x), clearly we substitute (x+∆x) in for the x value of the function (since you're saying that x = x+∆x).

Therefore, you'd have now for the difference quotient:

2(x+∆x)² - 8 - (2x²-8)
---------------------------------
∆x

By simplification and the using the foil method on (x+∆x)² you obtain:

2(x² + 2x∆x + (∆x)²) - 8 - 2x² + 8
-------------------------------------------------
∆x

Using the distributive property you obtain:

2x² + 4x∆x + 2(∆x)² - 8 -2x² + 8
-----------------------------------------------
∆x

The 2x² terms and the 8's cancel. This leaves you with:

4x∆x + 2(∆x)²
---------------------
∆x

Notice that you can factor out ∆x in both the numerator and denominator like so:

∆x(4x+2∆x)
-------------------
∆x

The ∆x's factored out will cancel out, leaving you with:

4x+2∆x

Some people use h in place of ∆x; so, you have two ways of writing the difference quotient:

1. 4x+2∆x

2. 4x+2h

There you go, I'm glad I could help.

f(x) = -8 + 2x^2
f(x) = 2x^2 - 8
f(x + h) = 2(x + h)^2 - 8
f(x + h) = 2(x + h)(x + h) - 8
f(x + h) = 2(x^2 + 2xh + h^2) - 8
f(x + h) = 2x^2 + 4xh + 2h^2 - 8

f'(x) = [f(x + h) - f(x)]/h
f'(x) = [2x^2 + 4xh + 2h^2 - 8 - (2x^2 - 8)]/h
f'(x) = [2x^2 + 4xh + 2h^2 - 8 - 2x^2 + 8]/h
f'(x) = [2x^2 - 2x^2 + 4xh + 2h^2 - 8 + 8]/h
f'(x) = [4xh + 2h^2]/h
f'(x) = 4x + 2h

you have troubles doing algebra...and you are in calculus??
1) f(x+h) = -8 + 2 [x+h]² = -8 + 2x² + 4 xh + 2 h²
2) f(x) =...........................-8 + 2x²
3) f(x+h) - f(x) = ..............................4 xh + 2 h² = h[4x + 2h]
4) divide 3) by h...................[ 4x + 2h]....finished

The differenxe quotient of a function, f(x), is this:

[ f(x + δx) - f(x) ] / δx

δx ≠ 0

f(x) = -8 + 2x²

f(x + δx)
= -8 + 2(x + δx)²
= -8 + 2x² + 4x(δx) + 2(δx)²

f(x + δx) - f(x)
= - 8 + 2x² + 4x(δx) + 2(δx)² - (-8 + 2x²)
= 4x(δx) + 2(δx)²
= δx(4x + 2δx)

[ f(x + δx) - f(x) ] / δx

= [ δx(4x + 2δx) ] / δx
= 4x + 2δx ; δx ≠ 0