Solved on Feb 12, 2024

Find the product and ratio of $f(x)=x^{1/2}$ and $g(x)=\sqrt[3]{2x}+1$ , and rationalize the denominator of the ratio.

STEP 1

Assumptions
1. The function $f(x)$ is defined as $f(x)=x^{\frac{1}{2}}$ .
2. The function $g(x)$ is defined as $g(x)=\sqrt[3]{2x}+1$ .
3. The operations to be performed are multiplication and division of the two functions.
4. When dividing, we will rationalize the denominator.

STEP 2

To find $f(x) \bullet g(x)$ , we multiply the two functions together.
$f(x) \bullet g(x) = f(x) \times g(x)$

STEP 3

Substitute the expressions for $f(x)$ and $g(x)$ into the multiplication.
$f(x) \bullet g(x) = x^{\frac{1}{2}} \times (\sqrt[3]{2x}+1)$

STEP 4

Distribute the multiplication over the addition in the parentheses.
$f(x) \bullet g(x) = x^{\frac{1}{2}} \times \sqrt[3]{2x} + x^{\frac{1}{2}} \times 1$

STEP 5

Simplify the multiplication by combining the exponents where possible.
$f(x) \bullet g(x) = (x^{\frac{1}{2}} \times x^{\frac{1}{3}}) \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

STEP 6

Use the property of exponents that states $a^{m} \times a^{n} = a^{m+n}$ to combine the exponents of $x$ .
$f(x) \bullet g(x) = x^{\frac{1}{2} + \frac{1}{3}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

STEP 7

Add the exponents of $x$ .
$f(x) \bullet g(x) = x^{\frac{3}{6} + \frac{2}{6}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$
$f(x) \bullet g(x) = x^{\frac{5}{6}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

STEP 8

Now we have the simplified form of $f(x) \bullet g(x)$ .
$f(x) \bullet g(x) = 2^{\frac{1}{3}}x^{\frac{5}{6}} + x^{\frac{1}{2}}$

STEP 9

Next, we find $\frac{f(x)}{g(x)}$ , which means dividing $f(x)$ by $g(x)$ .
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1}$

STEP 10

To rationalize the denominator, we need to multiply the numerator and the denominator by the conjugate of the denominator.
The conjugate of $\sqrt[3]{2x}+1$ is $\sqrt[3]{2x}-1$ .

STEP 11

Multiply the numerator and the denominator by the conjugate of the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1} \times \frac{\sqrt[3]{2x}-1}{\sqrt[3]{2x}-1}$

STEP 12

Perform the multiplication in the numerator and apply the difference of squares formula in the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{(\sqrt[3]{2x}+1)(\sqrt[3]{2x}-1)}$

STEP 13

Simplify the denominator using the difference of squares formula: $(a+b)(a-b)=a^2-b^2$ .
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{\sqrt[3]{(2x)^2}-1^2}$

STEP 14

Calculate the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{2^{\frac{2}{3}}x^{\frac{2}{3}}-1}$

STEP 15

Now we have the simplified form of $\frac{f(x)}{g(x)}$ with a rationalized denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{2^{\frac{2}{3}}x^{\frac{2}{3}}-1}$
a) The product $f(x) \bullet g(x)$ is $2^{\frac{1}{3}}x^{\frac{5}{6}} + x^{\frac{1}{2}}$ . b) The quotient $\frac{f(x)}{g(x)}$ with a rationalized denominator is $\frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{2^{\frac{2}{3}}x^{\frac{2}{3}}-1}$ .

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Find the product and ratio of $f(x)=x^{1/2}$ and $g(x)=\sqrt[3]{2x}+1$ , and rationalize the denominator of the ratio.

STEP 1

Assumptions
1. The function $f(x)$ is defined as $f(x)=x^{\frac{1}{2}}$ .
2. The function $g(x)$ is defined as $g(x)=\sqrt[3]{2x}+1$ .
3. The operations to be performed are multiplication and division of the two functions.
4. When dividing, we will rationalize the denominator.

STEP 2

To find $f(x) \bullet g(x)$ , we multiply the two functions together.
$f(x) \bullet g(x) = f(x) \times g(x)$

STEP 3

Substitute the expressions for $f(x)$ and $g(x)$ into the multiplication.
$f(x) \bullet g(x) = x^{\frac{1}{2}} \times (\sqrt[3]{2x}+1)$

STEP 4

Distribute the multiplication over the addition in the parentheses.
$f(x) \bullet g(x) = x^{\frac{1}{2}} \times \sqrt[3]{2x} + x^{\frac{1}{2}} \times 1$

STEP 5

Simplify the multiplication by combining the exponents where possible.
$f(x) \bullet g(x) = (x^{\frac{1}{2}} \times x^{\frac{1}{3}}) \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

STEP 6

Use the property of exponents that states $a^{m} \times a^{n} = a^{m+n}$ to combine the exponents of $x$ .
$f(x) \bullet g(x) = x^{\frac{1}{2} + \frac{1}{3}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

STEP 7

Add the exponents of $x$ .
$f(x) \bullet g(x) = x^{\frac{3}{6} + \frac{2}{6}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$
$f(x) \bullet g(x) = x^{\frac{5}{6}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

STEP 8

Now we have the simplified form of $f(x) \bullet g(x)$ .
$f(x) \bullet g(x) = 2^{\frac{1}{3}}x^{\frac{5}{6}} + x^{\frac{1}{2}}$

STEP 9

Next, we find $\frac{f(x)}{g(x)}$ , which means dividing $f(x)$ by $g(x)$ .
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1}$

STEP 10

To rationalize the denominator, we need to multiply the numerator and the denominator by the conjugate of the denominator.
The conjugate of $\sqrt[3]{2x}+1$ is $\sqrt[3]{2x}-1$ .

STEP 11

Multiply the numerator and the denominator by the conjugate of the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1} \times \frac{\sqrt[3]{2x}-1}{\sqrt[3]{2x}-1}$

STEP 12

Perform the multiplication in the numerator and apply the difference of squares formula in the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{(\sqrt[3]{2x}+1)(\sqrt[3]{2x}-1)}$

STEP 13

Simplify the denominator using the difference of squares formula: $(a+b)(a-b)=a^2-b^2$ .
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{\sqrt[3]{(2x)^2}-1^2}$

STEP 14

Calculate the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{2^{\frac{2}{3}}x^{\frac{2}{3}}-1}$

STEP 15

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Find the product and ratio of f(x)=x1/2f(x)=x^{1/2}f(x)=x1/2 and g(x)=2x3+1g(x)=\sqrt[3]{2x}+1g(x)=32x​+1, and rationalize the denominator of the ratio.

STEP 1

STEP 2

To find f(x)∙g(x)f(x) \bullet g(x)f(x)∙g(x), we multiply the two functions together.f(x)∙g(x)=f(x)×g(x)f(x) \bullet g(x) = f(x) \times g(x)f(x)∙g(x)=f(x)×g(x)

STEP 3

Substitute the expressions for f(x)f(x)f(x) and g(x)g(x)g(x) into the multiplication.f(x)∙g(x)=x12×(2x3+1)f(x) \bullet g(x) = x^{\frac{1}{2}} \times (\sqrt[3]{2x}+1)f(x)∙g(x)=x21​×(32x​+1)

STEP 4

Distribute the multiplication over the addition in the parentheses.f(x)∙g(x)=x12×2x3+x12×1f(x) \bullet g(x) = x^{\frac{1}{2}} \times \sqrt[3]{2x} + x^{\frac{1}{2}} \times 1f(x)∙g(x)=x21​×32x​+x21​×1

STEP 5

Simplify the multiplication by combining the exponents where possible.f(x)∙g(x)=(x12×x13)×213+x12f(x) \bullet g(x) = (x^{\frac{1}{2}} \times x^{\frac{1}{3}}) \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}f(x)∙g(x)=(x21​×x31​)×231​+x21​

STEP 6

Use the property of exponents that states am×an=am+na^{m} \times a^{n} = a^{m+n}am×an=am+n to combine the exponents of xxx.f(x)∙g(x)=x12+13×213+x12f(x) \bullet g(x) = x^{\frac{1}{2} + \frac{1}{3}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}f(x)∙g(x)=x21​+31​×231​+x21​

STEP 7

STEP 8

Now we have the simplified form of f(x)∙g(x)f(x) \bullet g(x)f(x)∙g(x).f(x)∙g(x)=213x56+x12f(x) \bullet g(x) = 2^{\frac{1}{3}}x^{\frac{5}{6}} + x^{\frac{1}{2}}f(x)∙g(x)=231​x65​+x21​

STEP 9

Next, we find f(x)g(x)\frac{f(x)}{g(x)}g(x)f(x)​, which means dividing f(x)f(x)f(x) by g(x)g(x)g(x).f(x)g(x)=x122x3+1\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1}g(x)f(x)​=32x​+1x21​​

STEP 10

To rationalize the denominator, we need to multiply the numerator and the denominator by the conjugate of the denominator.The conjugate of 2x3+1\sqrt[3]{2x}+132x​+1 is 2x3−1\sqrt[3]{2x}-132x​−1.

STEP 11

Multiply the numerator and the denominator by the conjugate of the denominator.f(x)g(x)=x122x3+1×2x3−12x3−1\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1} \times \frac{\sqrt[3]{2x}-1}{\sqrt[3]{2x}-1}g(x)f(x)​=32x​+1x21​​×32x​−132x​−1​

STEP 12

Perform the multiplication in the numerator and apply the difference of squares formula in the denominator.f(x)g(x)=x12(2x3−1)(2x3+1)(2x3−1)\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{(\sqrt[3]{2x}+1)(\sqrt[3]{2x}-1)}g(x)f(x)​=(32x​+1)(32x​−1)x21​(32x​−1)​

STEP 13

Simplify the denominator using the difference of squares formula: (a+b)(a−b)=a2−b2(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a2−b2.f(x)g(x)=x12(2x3−1)(2x)23−12\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{\sqrt[3]{(2x)^2}-1^2}g(x)f(x)​=3(2x)2​−12x21​(32x​−1)​

STEP 14

Calculate the denominator.f(x)g(x)=x12(2x3−1)223x23−1\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{2^{\frac{2}{3}}x^{\frac{2}{3}}-1}g(x)f(x)​=232​x32​−1x21​(32x​−1)​

STEP 15

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Find the product and ratio of $f(x)=x^{1/2}$ and $g(x)=\sqrt[3]{2x}+1$ , and rationalize the denominator of the ratio.

To find $f(x) \bullet g(x)$ , we multiply the two functions together.
$f(x) \bullet g(x) = f(x) \times g(x)$

Substitute the expressions for $f(x)$ and $g(x)$ into the multiplication.
$f(x) \bullet g(x) = x^{\frac{1}{2}} \times (\sqrt[3]{2x}+1)$

Distribute the multiplication over the addition in the parentheses.
$f(x) \bullet g(x) = x^{\frac{1}{2}} \times \sqrt[3]{2x} + x^{\frac{1}{2}} \times 1$

Simplify the multiplication by combining the exponents where possible.
$f(x) \bullet g(x) = (x^{\frac{1}{2}} \times x^{\frac{1}{3}}) \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

Use the property of exponents that states $a^{m} \times a^{n} = a^{m+n}$ to combine the exponents of $x$ .
$f(x) \bullet g(x) = x^{\frac{1}{2} + \frac{1}{3}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}$

Now we have the simplified form of $f(x) \bullet g(x)$ .
$f(x) \bullet g(x) = 2^{\frac{1}{3}}x^{\frac{5}{6}} + x^{\frac{1}{2}}$

Next, we find $\frac{f(x)}{g(x)}$ , which means dividing $f(x)$ by $g(x)$ .
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1}$

To rationalize the denominator, we need to multiply the numerator and the denominator by the conjugate of the denominator.
The conjugate of $\sqrt[3]{2x}+1$ is $\sqrt[3]{2x}-1$ .

Multiply the numerator and the denominator by the conjugate of the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}}{\sqrt[3]{2x}+1} \times \frac{\sqrt[3]{2x}-1}{\sqrt[3]{2x}-1}$

Perform the multiplication in the numerator and apply the difference of squares formula in the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{(\sqrt[3]{2x}+1)(\sqrt[3]{2x}-1)}$

Simplify the denominator using the difference of squares formula: $(a+b)(a-b)=a^2-b^2$ .
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{\sqrt[3]{(2x)^2}-1^2}$

Calculate the denominator.
$\frac{f(x)}{g(x)} = \frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{2^{\frac{2}{3}}x^{\frac{2}{3}}-1}$