Question 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.

Studdy Solution

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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