Answer
(a) $f(x) = \frac{x^2-1}{x^2+1}$
$f'(x) = \frac{4x}{(x^2+1)^2}$
$f''(x) = \frac{-12x^2+4}{(x^2+1)^3}$
(b) We can see a sketch of the three graphs.
$f'(x)$ 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
(a) $f(x) = \frac{x^2-1}{x^2+1}$
We can find $f'(x)$:
$f'(x) = \frac{(2x)(x^2+1)-(2x)(x^2-1)}{(x^2+1)^2}$
$f'(x) = \frac{2x^3+2x-2x^3+2x}{(x^2+1)^2}$
$f'(x) = \frac{4x}{(x^2+1)^2}$
We can find $f''(x)$:
$f''(x) = \frac{4(x^2+1)^2-2(x^2+1)(2x)(4x)}{(x^2+1)^4}$
$f''(x) = \frac{4(x^4+2x^2+1)-16x^4-16x^2}{(x^2+1)^4}$
$f''(x) = \frac{-12x^4-8x^2+4}{(x^2+1)^4}$
$f''(x) = \frac{(x^2+1)(-12x^2+4)}{(x^2+1)^4}$
$f''(x) = \frac{-12x^2+4}{(x^2+1)^3}$
(b) We can see a sketch of the three graphs.
$f'(x)$ 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.
