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. $8x_{\frac{1}{3}}$ and $(8x)_{\frac{1}{3}}$
c. $-2x$ and $(2x)^{-1}$

Explanation:

Step1: Choose a sub - problem (e.g., part a)

We will take part a: $\frac{1}{2}x$ and $x^{\frac{1}{2}}$. Let's pick a numerical value for $x$, say $x = 4$.

Step2: Calculate $\frac{1}{2}x$ when $x = 4$

Substitute $x = 4$ into $\frac{1}{2}x$. We get $\frac{1}{2}\times4=2$.

Step3: Calculate $x^{\frac{1}{2}}$ when $x = 4$

Recall that $x^{\frac{1}{2}}=\sqrt{x}$. So when $x = 4$, $x^{\frac{1}{2}}=\sqrt{4} = 2$? Wait, that's the same. Let's pick $x = 9$.

Step4: Recalculate for $x = 9$

For $\frac{1}{2}x$, we have $\frac{1}{2}\times9 = 4.5$. For $x^{\frac{1}{2}}$, we have $\sqrt{9}=3$. Now we can see the difference. The expression $\frac{1}{2}x$ means taking half of the number $x$, while $x^{\frac{1}{2}}$ means taking the square root of the number $x$.

(If we take part b: $8x^{\frac{1}{3}}$ and $(8x)^{\frac{1}{3}}$. Let $x = 1$. Then $8x^{\frac{1}{3}}=8\times1^{\frac{1}{3}} = 8$, and $(8x)^{\frac{1}{3}}=(8\times1)^{\frac{1}{3}}=\sqrt[3]{8}=2$. The first expression is 8 times the cube - root of $x$, the second is the cube - root of $8x$.

For part c: $- 2x$ and $(2x)^{-1}$. Let $x = 1$. Then $-2x=-2\times1=-2$, and $(2x)^{-1}=\frac{1}{2x}=\frac{1}{2\times1}=0.5$. The expression $-2x$ is - 2 times $x$, and $(2x)^{-1}$ is the reciprocal of $2x$.)

Answer:

Taking part a: When $x = 9$, $\frac{1}{2}x=4.5$ and $x^{\frac{1}{2}} = 3$. $\frac{1}{2}x$ represents half of $x$, $x^{\frac{1}{2}}$ represents the square root of $x$. (Or for part b: When $x = 1$, $8x^{\frac{1}{3}} = 8$, $(8x)^{\frac{1}{3}}=2$. $8x^{\frac{1}{3}}$ is 8 times the cube - root of $x$, $(8x)^{\frac{1}{3}}$ is the cube - root of $8x$. Or for part c: When $x = 1$, $-2x=-2$, $(2x)^{-1}=0.5$. $-2x$ is - 2 times $x$, $(2x)^{-1}$ is the reciprocal of $2x$.)