Find the Square Root End Point square root of 3x^5+7

$\sqrt{3 x^{5} + 7}$

Step 1

Find the domain for $\sqrt{3 x^{5} + 7}$ .

Set the radicand in $\sqrt{3 x^{5} + 7}$ greater than or equal to $0$ to find where the expression is defined.

$3 x^{5} + 7 \geq 0$

Solve for $x$ .

Subtract $7$ from both sides of the inequality.

$3 x^{5} \geq - 7$

Divide each term in $3 x^{5} \geq - 7$ by $3$ and simplify.

Divide each term in $3 x^{5} \geq - 7$ by $3$ .

$\frac{3 x^{5}}{3} \geq \frac{- 7}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{5}}{3} \geq \frac{- 7}{3}$

Divide $x^{5}$ by $1$ .

$x^{5} \geq \frac{- 7}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{5} \geq - \frac{7}{3}$

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt[5]{x^{5}} \geq \sqrt[5]{- \frac{7}{3}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$x \geq \sqrt[5]{- \frac{7}{3}}$

Simplify the right side.

Simplify $\sqrt[5]{- \frac{7}{3}}$ .

Rewrite $- \frac{7}{3}$ as ${({(- 1)}^{5})}^{5} \frac{7}{3}$ .

Rewrite $- 1$ as ${(- 1)}^{5}$ .

$x \geq \sqrt[5]{{(- 1)}^{5} \frac{7}{3}}$

Rewrite $- 1$ as ${(- 1)}^{5}$ .

$x \geq \sqrt[5]{{({(- 1)}^{5})}^{5} \frac{7}{3}}$

Pull terms out from under the radical.

$x \geq {(- 1)}^{5} \sqrt[5]{\frac{7}{3}}$

Raise $- 1$ to the power of $5$ .

$x \geq - \sqrt[5]{\frac{7}{3}}$

Rewrite $\sqrt[5]{\frac{7}{3}}$ as $\frac{\sqrt[5]{7}}{\sqrt[5]{3}}$ .

$x \geq - \frac{\sqrt[5]{7}}{\sqrt[5]{3}}$

Multiply $\frac{\sqrt[5]{7}}{\sqrt[5]{3}}$ by $\frac{{\sqrt[5]{3}}^{4}}{{\sqrt[5]{3}}^{4}}$ .

$x \geq - (\frac{\sqrt[5]{7}}{\sqrt[5]{3}} \cdot \frac{{\sqrt[5]{3}}^{4}}{{\sqrt[5]{3}}^{4}})$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[5]{7}}{\sqrt[5]{3}}$ by $\frac{{\sqrt[5]{3}}^{4}}{{\sqrt[5]{3}}^{4}}$ .

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{\sqrt[5]{3} {\sqrt[5]{3}}^{4}}$

Raise $\sqrt[5]{3}$ to the power of $1$ .

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{{\sqrt[5]{3}}^{1} {\sqrt[5]{3}}^{4}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{{\sqrt[5]{3}}^{1 + 4}}$

Add $1$ and $4$ .

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{{\sqrt[5]{3}}^{5}}$

Rewrite ${\sqrt[5]{3}}^{5}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{3}$ as $3^{\frac{1}{5}}$ .

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{{(3^{\frac{1}{5}})}^{5}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{3^{\frac{1}{5} \cdot 5}}$

Combine $\frac{1}{5}$ and $5$ .

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{3^{\frac{5}{5}}}$

Cancel the common factor of $5$ .

Cancel the common factor.

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{3^{\frac{5}{5}}}$

Rewrite the expression.

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{3^{1}}$

Evaluate the exponent.

$x \geq - \frac{\sqrt[5]{7} {\sqrt[5]{3}}^{4}}{3}$

Simplify the numerator.

Rewrite ${\sqrt[5]{3}}^{4}$ as $\sqrt[5]{3^{4}}$ .

$x \geq - \frac{\sqrt[5]{7} \sqrt[5]{3^{4}}}{3}$

Raise $3$ to the power of $4$ .

$x \geq - \frac{\sqrt[5]{7} \sqrt[5]{81}}{3}$

Simplify the numerator.

Combine using the product rule for radicals.

$x \geq - \frac{\sqrt[5]{7 \cdot 81}}{3}$

Multiply $7$ by $81$ .

$x \geq - \frac{\sqrt[5]{567}}{3}$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[- \frac{\sqrt[5]{567}}{3}, \infty)$

Set-Builder Notation:

${x | x \geq - \frac{\sqrt[5]{567}}{3}}$

Interval Notation:

$[- \frac{\sqrt[5]{567}}{3}, \infty)$

Set-Builder Notation:

${x | x \geq - \frac{\sqrt[5]{567}}{3}}$

Step 2

To find the square root end point, substitute the $x$ value $- \frac{\sqrt[5]{567}}{3}$ , which is the terminal value in the domain, into $\sqrt{3 x^{5} + 7}$ .

Replace the variable $x$ with $- \frac{\sqrt[5]{567}}{3}$ in the expression.

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 {(- \frac{\sqrt[5]{567}}{3})}^{5} + 7}$

Simplify the result.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{\sqrt[5]{567}}{3}$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 ({(- 1)}^{5} {(\frac{\sqrt[5]{567}}{3})}^{5}) + 7}$

Apply the product rule to $\frac{\sqrt[5]{567}}{3}$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 ({(- 1)}^{5} (\frac{{\sqrt[5]{567}}^{5}}{3^{5}})) + 7}$

Raise $- 1$ to the power of $5$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{{\sqrt[5]{567}}^{5}}{3^{5}}) + 7}$

Rewrite ${\sqrt[5]{567}}^{5}$ as $567$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{567}$ as $567^{\frac{1}{5}}$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{{(567^{\frac{1}{5}})}^{5}}{3^{5}}) + 7}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{567^{\frac{1}{5} \cdot 5}}{3^{5}}) + 7}$

Combine $\frac{1}{5}$ and $5$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{567^{\frac{5}{5}}}{3^{5}}) + 7}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{567^{\frac{5}{5}}}{3^{5}}) + 7}$

Rewrite the expression.

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{567}{3^{5}}) + 7}$

Evaluate the exponent.

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{567}{3^{5}}) + 7}$

Raise $3$ to the power of $5$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (- \frac{567}{243}) + 7}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{567}{243}$ into the numerator.

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (\frac{- 567}{243}) + 7}$

Factor $3$ out of $243$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (\frac{- 567}{3 (81)}) + 7}$

Cancel the common factor.

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{3 (\frac{- 567}{3 \cdot 81}) + 7}$

Rewrite the expression.

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{\frac{- 567}{81} + 7}$

Simplify the expression.

Divide $- 567$ by $81$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{- 7 + 7}$

Add $- 7$ and $7$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{0}$

Rewrite $0$ as $0^{2}$ .

$f (- \frac{\sqrt[5]{567}}{3}) = \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$f (- \frac{\sqrt[5]{567}}{3}) = 0$

The final answer is $0$ .

$y = 0$

Step 3

The square root end point is $(- \frac{\sqrt[5]{567}}{3}, 0)$ .

$(- \frac{\sqrt[5]{567}}{3}, 0)$

Step 4

Find the Square Root End Point square root of 3x^5+7

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