We are given the function $f(x) = x\sqrt{5-x}$. We need to find: (a) The domain of $f$. (b) The x and y-intercepts. (c) The first derivative $f'(x)$ and the second derivative $f''(x)$. (d) The critical points of $f$. (e) Where the curve is increasing and decreasing. (f) The points of inflection and the concavity of the curve. (g) The asymptotes. (h) Sketch the curve.

1. Problem Description

We are given the function $f(x) = x\sqrt{5-x}$. We need to find:

(a) The domain of $f$.

(b) The x and y-intercepts.

(c) The first derivative $f'(x)$ and the second derivative $f''(x)$.

(d) The critical points of $f$.

(e) Where the curve is increasing and decreasing.

(f) The points of inflection and the concavity of the curve.

(g) The asymptotes.

(h) Sketch the curve.

2. Solution Steps

(a) Domain of $f$:

The function $f(x) = x\sqrt{5-x}$ is defined when $5-x \ge 0$, which means $x \le 5$. Therefore, the domain of $f$ is $(-\infty, 5]$.

(b) Intercepts:

x-intercepts: $f(x) = 0$, so $x\sqrt{5-x} = 0$. Thus, $x=0$ or $\sqrt{5-x} = 0$, which gives $5-x=0$, so $x=5$. The x-intercepts are $x=0$ and $x=5$.

y-intercept: $f(0) = 0\sqrt{5-0} = 0$. The y-intercept is $y=0$.

(c) Derivatives:

$f(x) = x\sqrt{5-x} = x(5-x)^{1/2}$

$f'(x) = (1)(5-x)^{1/2} + x(\frac{1}{2}(5-x)^{-1/2}(-1))$

$f'(x) = \sqrt{5-x} - \frac{x}{2\sqrt{5-x}} = \frac{2(5-x)-x}{2\sqrt{5-x}} = \frac{10-2x-x}{2\sqrt{5-x}} = \frac{10-3x}{2\sqrt{5-x}}$

$f''(x) = \frac{(-3)(2\sqrt{5-x}) - (10-3x)(2(\frac{1}{2})(5-x)^{-1/2}(-1))}{4(5-x)}$

$f''(x) = \frac{-6\sqrt{5-x} + (10-3x)(5-x)^{-1/2}}{4(5-x)}$

$f''(x) = \frac{-6(5-x) + (10-3x)}{4(5-x)^{3/2}} = \frac{-30+6x+10-3x}{4(5-x)^{3/2}} = \frac{3x-20}{4(5-x)^{3/2}}$

(d) Critical points:

$f'(x) = \frac{10-3x}{2\sqrt{5-x}} = 0$ when $10-3x=0$, so $x = \frac{10}{3}$. Also, $f'(x)$ is undefined when $5-x = 0$, so $x=5$. Thus, the critical points are $x = \frac{10}{3}$ and $x=5$.

(e) Increasing/Decreasing:

$f'(x) = \frac{10-3x}{2\sqrt{5-x}}$

If $x < \frac{10}{3}$, then $10-3x > 0$, so $f'(x) > 0$. Thus, $f$ is increasing on $(-\infty, \frac{10}{3})$.

If $\frac{10}{3} < x < 5$, then $10-3x < 0$, so $f'(x) < 0$. Thus, $f$ is decreasing on $(\frac{10}{3}, 5)$.

At $x=5$, $f'(x)$ is undefined, and since the domain ends there, we consider $x=5$ to be the end of decreasing.

(f) Inflection points and Concavity:

$f''(x) = \frac{3x-20}{4(5-x)^{3/2}} = 0$ when $3x-20=0$, so $x = \frac{20}{3}$.

However, $\frac{20}{3} > 5$, so $x = \frac{20}{3}$ is not in the domain of $f$. Also, $f''(x)$ is undefined at $x=5$.

If $x < 5$, then $5-x > 0$, so $(5-x)^{3/2} > 0$.

If $x < 5$, then $3x-20 < 15-20 = -5 < 0$. Thus $f''(x) < 0$ on $(-\infty, 5)$.

Since the second derivative is always negative, the function is always concave down and there are no inflection points.

(g) Asymptotes:

Since the domain is $(-\infty, 5]$, there is no vertical asymptote at $x=5$. There are no other vertical asymptotes because the function is defined for all $x \le 5$. Since the domain does not extend to infinity, there are no horizontal asymptotes. There are no oblique asymptotes for the same reason. Thus, there are no asymptotes.

(h) Sketch: Key points: $(0,0)$, $(5,0)$. Critical point $(\frac{10}{3}, f(\frac{10}{3})) = (\frac{10}{3}, \frac{10}{3}\sqrt{5-\frac{10}{3}}) = (\frac{10}{3}, \frac{10}{3}\sqrt{\frac{5}{3}}) = (\frac{10}{3}, \frac{10\sqrt{15}}{9}) \approx (3.33, 4.3)$.

The function is increasing on $(-\infty, \frac{10}{3})$ and decreasing on $(\frac{10}{3}, 5)$. The function is always concave down.

3. Final Answer

(a) Domain: $(-\infty, 5]$

(b) Intercepts: x-intercepts: $x=0, 5$; y-intercept: $y=0$

(c) $f'(x) = \frac{10-3x}{2\sqrt{5-x}}$, $f''(x) = \frac{3x-20}{4(5-x)^{3/2}}$

(d) Critical points: $x=\frac{10}{3}, 5$

(e) Increasing: $(-\infty, \frac{10}{3})$; Decreasing: $(\frac{10}{3}, 5)$

(f) Inflection points: None; Concavity: Concave down on $(-\infty, 5)$

(g) Asymptotes: None

(h) (See explanation above for key points to use for the sketch.)