Question

1. the following pairs of expressions are not equivalent. how are their meanings different? give a numerical example to justify your response.

a. \\( \frac { 1 } { 2 } x \\) and \\( x ^ { \frac { 1 } { 2 } } \\)
b. \\( 8 x ^ { \frac { 1 } { 3 } } \\) and \\( ( 8 x ) ^ { \frac { 1 } { 3 } } \\)
c. \\( - 2 x \\) and \\( ( 2 x ) ^ { - 1 } \\)

Explanation:

a.

Step1: Analyze the meaning of $\frac{1}{2}x$ and $x^{\frac{1}{2}}$

$\frac{1}{2}x$ means half of \(x\). \(x^{\frac{1}{2}}=\sqrt{x}\), which means the non - negative square root of \(x\).

Step2: Provide a numerical example

Let \(x = 4\). Then \(\frac{1}{2}x=\frac{1}{2}\times4 = 2\), and \(x^{\frac{1}{2}}=\sqrt{4}=2\). Wait, this is wrong. Let's take \(x = 9\). \(\frac{1}{2}x=\frac{1}{2}\times9 = 4.5\), and \(x^{\frac{1}{2}}=\sqrt{9}=3\).

b.

Step1: Analyze the meaning of \(8x^{\frac{1}{3}}\) and \((8x)^{\frac{1}{3}}\)

\(8x^{\frac{1}{3}}=8\sqrt[3]{x}\), which means 8 times the cube root of \(x\). \((8x)^{\frac{1}{3}}=\sqrt[3]{8x}=2\sqrt[3]{x}\) (since \(\sqrt[3]{8}=2\)).

Step2: Provide a numerical example

Let \(x = 1\). \(8x^{\frac{1}{3}}=8\times\sqrt[3]{1}=8\), \((8x)^{\frac{1}{3}}=\sqrt[3]{8\times1}=2\).

c.

Step1: Analyze the meaning of \(- 2x\) and \((2x)^{-1}\)

\(-2x\) means \(-2\) times \(x\). \((2x)^{-1}=\frac{1}{2x}\) (\(x
eq0\)).

Step2: Provide a numerical example

Let \(x = 1\). \(-2x=-2\times1=-2\), \((2x)^{-1}=\frac{1}{2\times1}=\frac{1}{2}\).

Answer:

a. \(\frac{1}{2}x\) is half of \(x\), \(x^{\frac{1}{2}}\) is the square root of \(x\). For \(x = 9\), \(\frac{1}{2}x = 4.5\), \(x^{\frac{1}{2}}=3\).
b. \(8x^{\frac{1}{3}}\) is 8 times the cube root of \(x\), \((8x)^{\frac{1}{3}}\) is the cube root of \(8x\). For \(x = 1\), \(8x^{\frac{1}{3}}=8\), \((8x)^{\frac{1}{3}}=2\).
c. \(-2x\) is \(- 2\) times \(x\), \((2x)^{-1}\) is the reciprocal of \(2x\) (\(x
eq0\)). For \(x = 1\), \(-2x=-2\), \((2x)^{-1}=\frac{1}{2}\).