Question

1. describe the key features of the function ( f(x)=x^{-\frac{3}{2}} ).

Explanation:

Step1: Domain

For \(F(x)=x^{-\frac{3}{2}}=\frac{1}{x^{\frac{3}{2}}}=\frac{1}{\sqrt{x^{3}}}\), the expression under the square - root \(x^{3}>0\) (since the denominator cannot be zero). Solving \(x^{3}>0\) gives \(x > 0\). So the domain is \((0,\infty)\).

Step2: Range

As \(x>0\), when \(x\to0^{+}\), \(F(x)=x^{-\frac{3}{2}}=\frac{1}{x^{\frac{3}{2}}}\to\infty\). When \(x\to\infty\), \(F(x)=x^{-\frac{3}{2}}=\frac{1}{x^{\frac{3}{2}}}\to0\). So the range is \((0,\infty)\).

Step3: Continuity

The function \(y = x^{-\frac{3}{2}}\) is a power function. Since \(x>0\), and using the formula for power - functions \(y = x^{n}\) (where \(n=-\frac{3}{2}\)), and the rules of continuity for functions of the form \(y=\frac{1}{\sqrt{x^{3}}}\) (a composition of a root function and a rational function), it is continuous on its domain \((0,\infty)\).

Step4: Asymptotes

• Vertical Asymptote: As \(x\to0^{+}\), \(y = x^{-\frac{3}{2}}\to\infty\). So \(x = 0\) (the \(y\) - axis) is a vertical asymptote.
• Horizontal Asymptote: As \(x\to\infty\), \(y=x^{-\frac{3}{2}}=\frac{1}{x^{\frac{3}{2}}}\to0\). So \(y = 0\) (the \(x\) - axis) is a horizontal asymptote.

Step5: Monotonicity

Find the derivative using the power rule \(y=x^{n}\), \(y^\prime=nx^{n - 1}\). For \(y=x^{-\frac{3}{2}}\), \(y^\prime=-\frac{3}{2}x^{-\frac{3}{2}-1}=-\frac{3}{2}x^{-\frac{5}{2}}=-\frac{3}{2x^{\frac{5}{2}}}\). Since \(x>0\), \(y^\prime<0\) for all \(x\in(0,\infty)\). So the function is decreasing on the interval \((0,\infty)\).

Answer:

• Domain: \((0,\infty)\)
• Range: \((0,\infty)\)
• Continuity: Continuous on \((0,\infty)\)
• Vertical Asymptote: \(x = 0\)
• Horizontal Asymptote: \(y = 0\)
• Monotonicity: Decreasing on \((0,\infty)\)