Compute the first two derivative.
[tex]f(x) = x \sqrt{16 - x^2} = x (16 - x^2)^{1/2}[/tex]
[tex]f'(x) = x \dfrac{d}{dx} (16 - x^2)^{1/2} + (16 - x^2)^{1/2} \dfrac{d}{dx} x \\\\ ~~~~ = \dfrac x2 (16 - x^2)^{-1/2} \dfrac{d}{dx} (16-x^2) + (16 - x^2)^{1/2} \\\\ ~~~~ = \dfrac x2 (16 - x^2)^{-1/2} (-2x) + (16 - x^2)^{1/2} \\\\ ~~~~ = (16-x^2)^{-1/2} \left(-x^2 + (16 - x^2)\right) \\\\ ~~~~ = \dfrac{16 - 2x^2}{(16 - x^2)^{1/2}}[/tex]
[tex]f''(x) = \dfrac{(16-x^2)^{1/2} \frac d{dx} (16-2x^2) - (16-2x^2) \frac{d}{dx} (16-x^2)^{1/2}}{\left((16-x^2)^{1/2}\right)^2} \\\\ ~~~~ = \dfrac{(16-x^2)^{1/2} (-4x) - (8-x^2) (16 - x^2)^{-1/2} \frac{d}{dx} (16-x^2)}{16 - x^2} \\\\ ~~~~ = \dfrac{-4x (16-x^2) - (8-x^2) (-2x)}{(16 - x^2)^{3/2}} \\\\ ~~~~ = \dfrac{2x^3-48x}{(16 - x^2)^{3/2}}[/tex]
Take note of the domain of [tex]f(x)[/tex]. We must have [tex]16-x^2\ge0[/tex] for the square root to be defined, so
[tex]16 - x^2 \ge 0 \implies x^2 \le 16 \implies |x| \le 4[/tex]
and [tex]f(x)[/tex] exists only for [tex]-4 \le x \le 4[/tex].
Intercepts
Set [tex]x=0[/tex] to find the [tex]y[/tex]-intercepts. The only one is (0, 0), since
[tex]f(0) = 0 \sqrt{16-0^2} = 0[/tex]
The first intercept you listed is not a valid intercept. One or both coordinates must be 0.
Set [tex]f(x)=0[/tex] and solve for [tex]x[/tex] to find [tex]x[/tex]-intercepts. We already found (0, 0); there are two others at (-4, 0) and (4, 0).
[tex]f(x) = x \sqrt{16 - x^2} = 0 \\\\ x = 0 \text{ or } \sqrt{16 - x^2} = 0 \\\\ x = 0 \text{ or } 16 - x^2 = 0 \\\\ x = 0 \text{ or } x^2 = 16 \\\\ x = 0 \text{ or } x = \pm4[/tex]
The instructions say to list these in order from smallest to largest by [tex]x[/tex]-coordinate first, then by [tex]y[/tex]-coordinate. So the proper order of the intercepts would be (-4, 0), (0, 0), and (4, 0).
Relative minima/maxima
Find the critical points. We have [tex]f'(x) = 0[/tex] when
[tex]f'(x) = \dfrac{16 - 2x^2}{(16 - x^2)^{1/2}} = 0 \\\\ 16 - 2x^2 = 0 \\\\ x^2 = 8 \\\\ x = \pm2\sqrt2 \approx \pm 2.83[/tex]
and [tex]f'(x)[/tex] is undefined when
[tex](16 - x^2)^{1/2} = 0 \\\\ 16 - x^2 = 0 \\\\ x^2 = 16 \\\\ x = \pm4[/tex]
Check the sign of the second derivative at the first two critical points.
[tex]f''(-2\sqrt2) = 4 > 0 \implies \text{rel. min. at } f(-2\sqrt2) = -8[/tex]
[tex]f''(2\sqrt2) = -4 < 0 \implies \text{rel. max. at } f(2\sqrt2) = 8[/tex]
For posterity, we should also check the value of [tex]f(x)[/tex] at the endpoints of the domain.
[tex]f(-4) = f(4) = 0[/tex]
So [tex]f(x)[/tex] has a relative minimum at (-2.83, -8) and a relative maximum at (2.83, 8).
Inflection points
We have [tex]f''(x) = 0[/tex] for
[tex]f''(x) = \dfrac{2x^3-48x}{(16 - x^2)^{3/2}} = 0 \\\\ 2x^3 - 48x = 0 \\\\ 2x (x^2 - 24) = 0 \\\\ 2x = 0 \text{ or } x^2 - 24 = 0 \\\\ x = 0 \text{ or } x = \pm2\sqrt6\approx\pm4.90[/tex]
The two non-zero solutions fall outside the domain, so the only inflection point is (0, 0).
