Solved on Feb 12, 2024

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

STEP 1

Assumptions
1. The function f(x)f(x) is defined as f(x)=x12f(x)=x^{\frac{1}{2}}.
2. The function g(x)g(x) is defined as g(x)=2x3+1g(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)g(x)f(x) \bullet 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)

STEP 3

Substitute the expressions for f(x)f(x) and 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)

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 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}}

STEP 6

Use the property of exponents that states am×an=am+na^{m} \times a^{n} = a^{m+n} to combine the exponents of xx.
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}}

STEP 7

Add the exponents of xx.
f(x)g(x)=x36+26×213+x12f(x) \bullet g(x) = x^{\frac{3}{6} + \frac{2}{6}} \times 2^{\frac{1}{3}} + x^{\frac{1}{2}}
f(x)g(x)=x56×213+x12f(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)g(x)f(x) \bullet g(x).
f(x)g(x)=213x56+x12f(x) \bullet g(x) = 2^{\frac{1}{3}}x^{\frac{5}{6}} + x^{\frac{1}{2}}

STEP 9

Next, we find f(x)g(x)\frac{f(x)}{g(x)}, which means dividing f(x)f(x) by g(x)g(x).
f(x)g(x)=x122x3+1\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 2x3+1\sqrt[3]{2x}+1 is 2x31\sqrt[3]{2x}-1.

STEP 11

Multiply the numerator and the denominator by the conjugate of the denominator.
f(x)g(x)=x122x3+1×2x312x31\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.
f(x)g(x)=x12(2x31)(2x3+1)(2x31)\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)(ab)=a2b2(a+b)(a-b)=a^2-b^2.
f(x)g(x)=x12(2x31)(2x)2312\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.
f(x)g(x)=x12(2x31)223x231\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 f(x)g(x)\frac{f(x)}{g(x)} with a rationalized denominator.
f(x)g(x)=x12(2x31)223x231\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)g(x)f(x) \bullet g(x) is 213x56+x122^{\frac{1}{3}}x^{\frac{5}{6}} + x^{\frac{1}{2}}. b) The quotient f(x)g(x)\frac{f(x)}{g(x)} with a rationalized denominator is x12(2x31)223x231\frac{x^{\frac{1}{2}}(\sqrt[3]{2x}-1)}{2^{\frac{2}{3}}x^{\frac{2}{3}}-1}.

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