The natural domain is all real numbers except $-5$ and $2$, which make the denominator equal to $0$. These are singularities of $B$. Otherwise, $B$ is continuous. There are no discontinuities.

Domain = $(-\infty , -5) \cup (-5, 2) \cup (2, \infty )$

The graph will help us with the range, so we’ll wait on that.

$B$ has zeros at $\answer {-3}$ and $\answer {1}$. These make the numerator equal to $0$. These are represented on the graph with $(-3,0)$ and $(1,0)$

The graph will have vertical asymptotes $f=-5$ and $f=-2$. $B$ will change sign over $f=-5$, since the multiplicity is odd even . $B$ will not change sign over $f=2$, since the multiplicity is odd even .

$B$ is a rational function and the degree of the denominator is greater than less than the degree of the numerator. Therefore,

$$\lim _{f \to -\infty } B(f) = \answer {0} \, \text { and } \, \lim _{f \to \infty } B(f) = \answer {0}$$

The graph has a horizontal asymptote at $y=\answer {0}$.

From this, we can sketch a graph of $y = B(f)$.

Both zeros have odd multiplicity, therefore the graph must go below the $f$-axis between them. There must be a local minimum. From the graph we can estimate that the critical number is approximately $0.5$. $B(0.5) = \frac {14}{99} \approx -0.1414$ is a local minimum. With some Calculus tools, we may be able to get an exact value.

$B$ has no global maximum or minimum.

$B$ has no local maximum.

• $B$ decreases on $(-\infty , -5)$.
• $B$ decreases on $(-5, 0.5)$.
• $B$ increases on $(0.5, 2)$.
• $B$ decreases on $(2, \infty )$.

The graph makes it evident that the range is all real numbers.

The domain is restricted by the square root to be $\left ( -\infty , \answer {9} \right ]$.

The formula is a product, therefore the zeros come from each factor.

The factor of $w$ gives a zero of $\answer {0}$.

The factor of $\sqrt {9-w}$ gives a zero of $\answer {9}$.

Since $\sqrt {9-w}$ is always nonnegative, the sign of $K$ is the same as the sign of $w$.

• $K$ is negative on $\left ( \answer {-\infty },\answer {0} \right )$
• $K$ is positive on $\left ( \answer {0}, \answer {\infty } \right )$

From this, we can sketch a graph of $y = K(w)$.

Our sketch suggest a global maximum between $0$ and $9$.

If we had the derivative, then we could identify this critical number exactly. Currently, we’ll need a graph to approximate values.

• $K$ has a local and global maximum of $10.39$ at $6$.
• $K$ has a local minimum of $0$ at $9$.
• $K$ has no global minimum.

• $K$ is increasing on $(-\infty , 6]$.
• $K$ is decreasing on $[6, 9]$.

Finally, the range is approximately $(-\infty , 10.39]$.

The graph has no vertical asymptotes, $\lim \limits _{w \to -\infty } K(w) = -\infty .$