Slope: The Derivative of a Function at a Point

[morrowcosom]

Homework Statement

We are calculating the slope of the function f(x) = x^3 /3 - 2x at x = 2.
For the function f(x) = x^3 /3 - 2x, we now know:

f(2) = -4/3
f(2+h) = (8 + 12h + 6h^2 + h^3)/3 - (4 + 2h)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
{f(2+h) - f(2)}/h=
--------------------------------------------------------------------------------

Homework Equations

The Attempt at a Solution

I tried to simplify the f(2+h) equation by turning it into
{(4+10h+6h^2+h^3)/3 +4/3}/h (I notice here that the 4/3's seem to have no opposite signs :()
{(10h+6h^2+h^3)/3}/h
{h(10+6h+h^2)/3}/h
=(10+6h+h^2)/3

What did I do wrong? I think I screwed up on the blending in the - (4 + 2h)

 

[Mark44]

Homework Statement

We are calculating the slope of the function f(x) = x^3 /3 - 2x at x = 2.
For the function f(x) = x^3 /3 - 2x, we now know:

f(2) = -4/3
f(2+h) = (8 + 12h + 6h^2 + h^3)/3 - (4 + 2h)

Simplify the above to 8/3 + 4h + 2h^2 + (1/3)h^3 - 4 - 2h = -4/3 + 2h + 2h^2 + (1/3)h^3
Then subtract f(2).

Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
{f(2+h) - f(2)}/h=
--------------------------------------------------------------------------------

Homework Equations

The Attempt at a Solution

I tried to simplify the f(2+h) equation by turning it into
{(4+10h+6h^2+h^3)/3 +4/3}/h (I notice here that the 4/3's seem to have no opposite signs :()
{(10h+6h^2+h^3)/3}/h
{h(10+6h+h^2)/3}/h
=(10+6h+h^2)/3

What did I do wrong? I think I screwed up on the blending in the - (4 + 2h)