Answer
$f(x) = 3x^2+2x+1$
$f'(x) = 6x+2$
$f''(x) = 6$
We can see a sketch of the three graphs on the same screen.
Work Step by Step
$f(x) = 3x^2+2x+1$
We can find $f'(x)$:
$f'(x) = \lim\limits_{h \to 0}\frac{[3(x+h)^2+2(x+h)+1]-(3x^2+2x+1)}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{(3x^2+6xh+3h^2+2x+2h+1)-(3x^2+2x+1)}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{6xh+3h^2+2h}{h}$
$f'(x) = \lim\limits_{h \to 0} (6x+3h+2)$
$f'(x) = 6x+0+2$
$f'(x) = 6x+2$
We can find $f''(x)$:
$f''(x) = \lim\limits_{h \to 0}\frac{[6(x+h)+2]-(6x+2)}{h}$
$f''(x) = \lim\limits_{h \to 0}\frac{6h}{h}$
$f''(x) = \lim\limits_{h \to 0} 6$
$f''(x) = 6$
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 a constant slope of 6.
