the function, f(x)=15√x, is given. determine when the function is positive, negative, increasing, or…

Explanation:

Step1: Analyze when function is positive

The square - root function $\sqrt{x}$ is defined for $x\geq0$. Since $15>0$ and $\sqrt{x}\geq0$ for $x\geq0$, $f(x) = 15\sqrt{x}>0$ when $x>0$.

Step2: Analyze when function is negative

Since $\sqrt{x}\geq0$ for all $x$ in its domain ($x\geq0$) and $15>0$, $f(x)=15\sqrt{x}\geq0$ for all $x$ in its domain. So, there is no $x$ for which $f(x)<0$.

Step3: Analyze increasing or decreasing

The derivative of $y = 15\sqrt{x}=15x^{\frac{1}{2}}$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $y^\prime=\frac{15}{2\sqrt{x}}>0$ for $x > 0$. So the function is increasing for $x>0$.

Step4: Analyze end - behavior

As $x\rightarrow+\infty$, $\sqrt{x}\rightarrow+\infty$, and since $f(x)=15\sqrt{x}$, then $y\rightarrow+\infty$ as $x\rightarrow+\infty$.

Answer:

Positive: $x>0$ Negative: None Increasing: $x > 0$ Decreasing: None End behavior: $y\rightarrow+\infty$ as $x\rightarrow+\infty$