use a graph to (a) find the range, (b) give the largest open interval over which the function is increasing…

Explanation:

Step1: Determine domain of the function

For $f(x)=\sqrt{9 + 2x}$, the expression under the square - root must be non - negative. So, $9+2x\geq0$, which gives $x\geq-\frac{9}{2}$.

Step2: Analyze the range

Since the square - root function $\sqrt{u}$ has a range of $y\geq0$ when $u\geq0$. Here $u = 9 + 2x\geq0$, so the range of $f(x)=\sqrt{9 + 2x}$ is $y\geq0$.

Step3: Analyze increasing and decreasing intervals

The derivative of $y=\sqrt{9 + 2x}=(9 + 2x)^{\frac{1}{2}}$ using the chain rule: $y^\prime=\frac{1}{2}(9 + 2x)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{9 + 2x}}$. Since $y^\prime>0$ for all $x>-\frac{9}{2}$, the function is increasing on the interval $(-\frac{9}{2},\infty)$ and there is no interval where it is decreasing.

Step4: Solve $f(x)=0$

Set $\sqrt{9 + 2x}=0$. Squaring both sides gives $9 + 2x = 0$, so $x=-\frac{9}{2}$.

Answer:

a. Range: $[0,\infty)$ b. Increasing interval: $(-\frac{9}{2},\infty)$ c. Decreasing interval: None d. Solution of $f(x) = 0$: $x=-\frac{9}{2}$