Answer
$f(x) = x^3-3x$
$f'(x) = 3x^2-3$
$f''(x) = 6x$
We can see a sketch of the three graphs on the same screen.
$f'(x)$ seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope.
$f''(x)$ seems reasonable since $f''(x)$ has negative values when the slope of $f'(x)$ has a negative slope, $f''(x)$ is 0 when the slope of $f'(x)$ is 0, and $f''(x)$ has positive values when the slope of $f'(x)$ has a positive slope.
Work Step by Step
$f(x) = x^3-3x$
We can find $f'(x)$:
$f'(x) = \lim\limits_{h \to 0}\frac{[(x+h)^3-3(x+h)]-(x^3-3x)}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{(x^3+3x^2h+3xh^2+h^3-3x-3h)-(x^3-3x)}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{3x^2h+3xh^2+h^3-3h}{h}$
$f'(x) = \lim\limits_{h \to 0}(3x^2+3xh+h^2-3)$
$f'(x) = 3x^2+0+0-3$
$f'(x) = 3x^2-3$
We can find $f''(x)$:
$f''(x) = \lim\limits_{h \to 0}\frac{[3(x+h)^2-3]-(3x^2-3)}{h}$
$f''(x) = \lim\limits_{h \to 0}\frac{(3x^2+6xh+3h^2-3)-(3x^2-3)}{h}$
$f''(x) = \lim\limits_{h \to 0}\frac{6xh+3h^2}{h}$
$f''(x) = \lim\limits_{h \to 0}(6x+3h)$
$f''(x) = 6x+0$
$f''(x) = 6x$
We can see a sketch of the three graphs on the same screen.
$f'(x)$ seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope.
$f''(x)$ seems reasonable since $f''(x)$ has negative values when the slope of $f'(x)$ has a negative slope, $f''(x)$ is 0 when the slope of $f'(x)$ is 0, and $f''(x)$ has positive values when the slope of $f'(x)$ has a positive slope.
