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

Question

Luke Alvarez · Accepted Answer

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# 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#

Aurora Alvarado · Answer

undefined To find $ f'(x) $ using the limit definition for the function $ f(x) = -3x^3 + 9x + 4 $, follow these steps:

1. Write down the limit definition of the derivative: 
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

2. Substitute the given function $ f(x) = -3x^3 + 9x + 4 $ into the limit definition.

3. Expand and simplify the expression.

4. Take the limit as $ h $ approaches 0.

5. Simplify the expression further if possible to obtain $ f'(x) $.