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

1 Answer
Douglas K.
Jan 9, 2017

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#

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