Calculus Examples

$f (x) = \sqrt{2 x^{2} - x - 1}$

Step 1

Find the domain to determine if the expression is continuous.

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Step 1.1

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

$2 x^{2} - x - 1 \geq 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Convert the inequality to an equation.

$2 x^{2} - x - 1 = 0$

Step 1.2.2

Factor by grouping.

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Step 1.2.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

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Step 1.2.2.1.1

Factor $- 1$ out of $- x$ .

$2 x^{2} - (x) - 1 = 0$

Step 1.2.2.1.2

Rewrite $- 1$ as $1$ plus $- 2$

$2 x^{2} + (1 - 2) x - 1 = 0$

Step 1.2.2.1.3

Apply the distributive property.

$2 x^{2} + 1 x - 2 x - 1 = 0$

Step 1.2.2.1.4

Multiply $x$ by $1$ .

$2 x^{2} + x - 2 x - 1 = 0$

Step 1.2.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.2.1

Group the first two terms and the last two terms.

$(2 x^{2} + x) - 2 x - 1 = 0$

Step 1.2.2.2.2

Factor out the greatest common factor (GCF) from each group.

$x (2 x + 1) - (2 x + 1) = 0$

Step 1.2.2.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$(2 x + 1) (x - 1) = 0$

Step 1.2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x + 1 = 0$

$x - 1 = 0$

Step 1.2.4

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 1.2.4.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 1.2.4.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 1.2.4.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 1.2.4.2.2

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

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Step 1.2.4.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

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

Step 1.2.4.2.2.2

Simplify the left side.

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Step 1.2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.4.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.4.2.2.3

Simplify the right side.

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Step 1.2.4.2.2.3.1

Move the negative in front of the fraction.

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

Step 1.2.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 1.2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.2.6

The final solution is all the values that make $(2 x + 1) (x - 1) = 0$ true.

$x = - \frac{1}{2}, 1$

Step 1.2.7

Use each root to create test intervals.

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

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

$x > 1$

Step 1.2.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.2.8.1

Test a value on the interval $x < - \frac{1}{2}$ to see if it makes the inequality true.

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Step 1.2.8.1.1

Choose a value on the interval $x < - \frac{1}{2}$ and see if this value makes the original inequality true.

$x = - 3$

Step 1.2.8.1.2

Replace $x$ with $- 3$ in the original inequality.

$2 {(- 3)}^{2} - (- 3) - 1 \geq 0$

Step 1.2.8.1.3

The left side $20$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.2

Test a value on the interval $- \frac{1}{2} < x < 1$ to see if it makes the inequality true.

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Step 1.2.8.2.1

Choose a value on the interval $- \frac{1}{2} < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 1.2.8.2.2

Replace $x$ with $0$ in the original inequality.

$2 {(0)}^{2} - (0) - 1 \geq 0$

Step 1.2.8.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.8.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 1.2.8.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 1.2.8.3.2

Replace $x$ with $4$ in the original inequality.

$2 {(4)}^{2} - (4) - 1 \geq 0$

Step 1.2.8.3.3

The left side $27$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - \frac{1}{2}$ True

$- \frac{1}{2} < x < 1$ False

$x > 1$ True

$x < - \frac{1}{2}$ True

$- \frac{1}{2} < x < 1$ False

$x > 1$ True

Step 1.2.9

The solution consists of all of the true intervals.

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

Step 1.3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \leq - \frac{1}{2}, x \geq 1}$

Interval Notation:

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

Set-Builder Notation:

${x | x \leq - \frac{1}{2}, x \geq 1}$

Step 2

Since the domain is not all real numbers, $\sqrt{2 x^{2} - x - 1}$ is not continuous over all real numbers.

Not continuous

Step 3