Answer
(a) $f'(0) = 0$
$f'(\frac{1}{2}) = 0.75$
$f'(1) = 3$
$f'(2) = 12$
$f'(3) = 27$
(b) $f'(-\frac{1}{2}) = 0.75$
$f'(-1) = 3$
$f'(-2) = 12$
$f'(-3) = 27$
(c) We can see a sketch of the graph below.
(d) We could guess that $f'(x) = 3x^2$
(e) $f'(x) = 3x^2$
Work Step by Step
(a) $f(x) = x^3$
When we use the zoom function on a graphing device, we can estimate the following values for the slope at various points on the graph of $f(x)$:
$f'(0) = 0$
$f'(\frac{1}{2}) = 0.75$
$f'(1) = 3$
$f'(2) = 12$
$f'(3) = 27$
(b) We can use symmetry to deduce the following values:
$f'(-\frac{1}{2}) = 0.75$
$f'(-1) = 3$
$f'(-2) = 12$
$f'(-3) = 27$
(c) We can see a sketch of the graph below.
(d) The graph is a parabola that looks like the graph of $3x^2$
We could guess that $f'(x) = 3x^2$
(e) We can use the definition of derivative to find $f'(x)$:
$f'(x) = \lim\limits_{h \to 0}\frac{(x+h)^3-x^3}{h}$
$f'(x) =\lim\limits_{h \to 0}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}$
$f'(x) =\lim\limits_{h \to 0}\frac{3x^2h+3xh^2+h^3}{h}$
$f'(x) =\lim\limits_{h \to 0}(3x^2+3xh+h^2)$
$f'(x) =3x^2+0+0$
$f'(x) = 3x^2$
