Answer
$f(x) = 2x^2-x^3$
$f'(x) = 4x-3x^2$
$f''(x) = 4-6x$
$f'''(x) = -6$
$f^{(4)}(x) = 0$
The graphs are consistent with the geometric interpretations of these derivatives. The graph of each derivative has positive values when the slope of the previous function is positive, an x-intercept when the slope is zero, and negative values when the slope is negative.
Work Step by Step
$f(x) = 2x^2-x^3$
We can find $f'(x)$:
$f'(x) = \lim\limits_{h \to 0}\frac{[2(x+h)^2-(x+h)^3]-(2x^2-x^3)}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{(2x^2+4xh+2h^2-x^3-3x^2h-3xh^2-h^3)-(2x^2-x^3)}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{4xh+2h^2-3x^2h-3xh^2-h^3}{h}$
$f'(x) = \lim\limits_{h \to 0}(4x+2h-3x^2-3xh-h^2)$
$f'(x) = 4x+0-3x^2-0-0$
$f'(x) = 4x-3x^2$
We can find $f''(x)$:
$f''(x) = \lim\limits_{h \to 0}\frac{[4(x+h)-3(x+h)^2]-(4x-3x^2)}{h}$
$f''(x) = \lim\limits_{h \to 0}\frac{(4x+4h-3x^2-6xh-3h^2)-(4x-3x^2)}{h}$
$f''(x) = \lim\limits_{h \to 0}\frac{4h-6xh-3h^2}{h}$
$f''(x) = \lim\limits_{h \to 0}(4-6x-3h)$
$f''(x) = 4-6x-0$
$f''(x) = 4-6x$
We can find $f'''(x)$:
$f'''(x) = \lim\limits_{h \to 0}\frac{[4-6(x+h)]-(4-6x)}{h}$
$f'''(x) = \lim\limits_{h \to 0}\frac{(4-6x-6h)-(4-6x)}{h}$
$f'''(x) = \lim\limits_{h \to 0}\frac{-6h}{h}$
$f'''(x) = \lim\limits_{h \to 0}~-6$
$f'''(x) = -6$
We can find $f^{(4)}(x)$:
$f^{(4)}(x) = \lim\limits_{h \to 0}\frac{(-6)-(-6)}{h} = 0$
The graphs are consistent with the geometric interpretations of these derivatives. The graph of each derivative has positive values when the slope of the previous function is positive, an x-intercept when the slope is zero, and negative values when the slope is negative.
