Graph 4- square root of 2x+1<1

$4 - \sqrt{2 x + 1} < 1$

Step 1

Move all terms not containing $- \sqrt{2 x + 1}$ to the right side of the inequality.

Subtract $4$ from both sides of the inequality.

$- \sqrt{2 x + 1} < 1 - 4$

Subtract $4$ from $1$ .

$- \sqrt{2 x + 1} < - 3$

Step 2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${(- \sqrt{2 x + 1})}^{2} < {(- 3)}^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 1}$ as ${(2 x + 1)}^{\frac{1}{2}}$ .

${(- {(2 x + 1)}^{\frac{1}{2}})}^{2} < {(- 3)}^{2}$

Simplify the left side.

Simplify ${(- {(2 x + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(2 x + 1)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(2 x + 1)}^{\frac{1}{2}})}^{2} < {(- 3)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(2 x + 1)}^{\frac{1}{2}})}^{2} < {(- 3)}^{2}$

Multiply ${({(2 x + 1)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(2 x + 1)}^{\frac{1}{2}})}^{2} < {(- 3)}^{2}$

Multiply the exponents in ${({(2 x + 1)}^{\frac{1}{2}})}^{2}$ .

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

${(2 x + 1)}^{\frac{1}{2} \cdot 2} < {(- 3)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x + 1)}^{\frac{1}{2} \cdot 2} < {(- 3)}^{2}$

Rewrite the expression.

${(2 x + 1)}^{1} < {(- 3)}^{2}$

Simplify.

$2 x + 1 < {(- 3)}^{2}$

Simplify the right side.

Raise $- 3$ to the power of $2$ .

$2 x + 1 < 9$

Step 4

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $1$ from both sides of the inequality.

$2 x < 9 - 1$

Subtract $1$ from $9$ .

$2 x < 8$

Divide each term in $2 x < 8$ by $2$ and simplify.

Divide each term in $2 x < 8$ by $2$ .

$\frac{2 x}{2} < \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{8}{2}$

Divide $x$ by $1$ .

$x < \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$x < 4$

Step 5

Find the domain of $3 - \sqrt{2 x + 1}$ .

Set the radicand in $\sqrt{2 x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$2 x + 1 \geq 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$2 x \geq - 1$

Divide each term in $2 x \geq - 1$ by $2$ and simplify.

Divide each term in $2 x \geq - 1$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 1}{2}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{2}$

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

$[- \frac{1}{2}, \infty)$

Step 6

The solution consists of all of the true intervals.

$x > 4$

Step 7

Graph 4- square root of 2x+1<1

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