f(x) = -x^3 + 3x^2 -2 at x=1?

Question

A) For the following function find the value of the derivative at the specific point given using:
- the definition of the derivative
- the sum rule for derivatives
Show that both methods lead to the same results
f(x) = -x^3 + 3x^2 -2 at x=1

Answer 1

f(x) = -x^3 + 3x^2 -2

Definition of derivative:
lim: (f(x + h) - f(x)) / h
h->0

f(x + h) = -(x+h)^3 + 3(x+h)^2 - 2
f(x) = -x^3 + 3x^2 - 2

Expand the parenthesis, then simplify:
f'(x) = (-(x^3 + 3hx^2 + 3h^2x + h^3) + 3(x^2 + 2hx + h^2) - 2 + (x^3 + 3x^2 - 2)) / h
f'(x) = (-x^3 + x^3 - 3hx^2 - 3x^2 + 3x^2 - 3h^2x + 6hx - h^3 + 3h^2 - 2 + 2) / h
f'(x) = (-3hx^2 - 3h^2x + 6hx - h^3 + 3h^2) / h

Cancel out h:
f'(x) = -3x^2 - 3hx + 6x - h^2 + 3h

Since the limit is of h-> 0, every term left with an h becomes 0.
f'(x) = -3x^2 + 6x

Using the sum and power rules:
Power rule: for f(x) = an^b, f'(x) = (ba)n^(b-1)
Sum rule: for f(x) = g(x) + h(x), f'(x) = g'(x) + h'(x).

f(x) = -x^3 + 3x^2 -2
f'(x) = -3x^2 + 6x

at x=1:
f'(1) = -3(1)^2 + 6(1) = -3 + 6 = 3.