Question

How do you find f'(x) using the limit definition given `f(x) = -3 x^3 + 9 x + 4`?

Answer 1

Use `f'(x) = lim_(hto0)(f(x+h) - f(x))/h`
Write the simplest form of f(x+h)
Subtract f(x) from that.
A common factor of `h/h` will cancel.
Let `h to 0`

Explanation:

Use `f'(x) = lim_(hto0)(f(x+h) - f(x))/h`

Given: `f(x) = -3x^3 + 9x + 4`

Then write the expression for `f(x + h)`

`f(x+h) = -3(x+h)^3 + 9(x + h) + 4`

`f(x+h) = -3(x+h)(x^2 + 2hx + h^2) + 9(x + h) + 4`

`f(x+h) = -3(x^3 + 2hx^2 + h^2x + hx^2 + 2h^2x + h^2) + 9(x + h) + 4`

`f(x+h) = -3(x^3 + 3hx^2 + 3h^2x + h^2) + 9(x + h) + 4`

`f(x+h) = -3x^3 - 9hx^2 - 9h^2x - 9h^2 + 9x + 9h + 4`

The above is the simplest form of `f(x + h)`

Use that form to simplify the numerator:

`f(x+h) - f(x) = -3x^2 - 9hx^2 - 9h^2x - 9h^2 + 9x + 9h + 4 + 3x^3 - 9x - 4`

`f(x+h) - f(x) = -9hx^2 - 9h^2x - 9h^2 + 9h`

Remove a common factor, h:

`f(x+h) - f(x) = h(-9x^2 - 9hx - 9h + 9)`

Substitute the simplified numerator into the limit:

`f'(x) = lim_(hto0)(h(-9x^2 - 9hx - 9h + 9))/h`

`h/h` becomes 1:

`f'(x) = lim_(hto0)-9x^2 - 9hx - 9h + 9`

Let `h to 0`

`f'(x) = -9x^2 + 9`