Question

Calculus HW , will someone please teach me how to do this problem. I'm struggling, thanks! 10 Points

Answer 1

Compute the first two derivative.

`f(x) = x √(16 - x^2) = x (16 - x^2)^(1/2)`

`f'(x) = x (d)/(dx) (16 - x^2)^(1/2) + (16 - x^2)^(1/2) (d)/(dx) x \\\\ ~~~~ = \frac x2 (16 - x^2)^(-1/2) (d)/(dx) (16-x^2) + (16 - x^2)^(1/2) \\\\ ~~~~ = \frac x2 (16 - x^2)^(-1/2) (-2x) + (16 - x^2)^(1/2) \\\\ ~~~~ = (16-x^2)^(-1/2) \left(-x^2 + (16 - x^2)\right) \\\\ ~~~~ = (16 - 2x^2)/((16 - x^2)^(1/2))`

`f''(x) = \frac{(16-x^2)^(1/2) \frac d{dx} (16-2x^2) - (16-2x^2) (d)/(dx) (16-x^2)^(1/2)}{\left((16-x^2)^(1/2)\right)^2} \\\\ ~~~~ = ((16-x^2)^(1/2) (-4x) - (8-x^2) (16 - x^2)^(-1/2) (d)/(dx) (16-x^2))/(16 - x^2) \\\\ ~~~~ = (-4x (16-x^2) - (8-x^2) (-2x))/((16 - x^2)^(3/2)) \\\\ ~~~~ = (2x^3-48x)/((16 - x^2)^(3/2))`

Take note of the domain of
`f(x)`. We must have
`16-x^2\ge0` for the square root to be defined, so

`16 - x^2 \ge 0 \implies x^2 \le 16 \implies |x| \le 4`

and
`f(x)` exists only for
`-4 \le x \le 4`.

Intercepts

Set
`x=0` to find the
`y`-intercepts. The only one is (0, 0), since

`f(0) = 0 √(16-0^2) = 0`

The first intercept you listed is not a valid intercept. One or both coordinates must be 0.

Set
`f(x)=0` and solve for
`x` to find
`x`-intercepts. We already found (0, 0); there are two others at (-4, 0) and (4, 0).

`f(x) = x √(16 - x^2) = 0 \\\\ x = 0 \text{ or } √(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`

The instructions say to list these in order from smallest to largest by
`x`-coordinate first, then by
`y`-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
`f'(x) = 0` when

`f'(x) = (16 - 2x^2)/((16 - x^2)^(1/2)) = 0 \\\\ 16 - 2x^2 = 0 \\\\ x^2 = 8 \\\\ x = \pm2\sqrt2 \approx \pm 2.83`

and
`f'(x)` is undefined when

`(16 - x^2)^(1/2) = 0 \\\\ 16 - x^2 = 0 \\\\ x^2 = 16 \\\\ x = \pm4`

Check the sign of the second derivative at the first two critical points.

`f''(-2\sqrt2) = 4 > 0 \implies \text{rel. min. at }  f(-2\sqrt2) = -8`

`f''(2\sqrt2) = -4 < 0 \implies \text{rel. max. at } f(2\sqrt2) = 8`

For posterity, we should also check the value of
`f(x)` at the endpoints of the domain.

`f(-4) = f(4) = 0`

So
`f(x)` has a relative minimum at (-2.83, -8) and a relative maximum at (2.83, 8).

Inflection points

We have
`f''(x) = 0` for

`f''(x) = (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`

The two non-zero solutions fall outside the domain, so the only inflection point is (0, 0).