Calculus Examples

$f (x) = 2 x^{2} + 3 x + 1$

Step 1

Write $f (x) = 2 x^{2} + 3 x + 1$ as an equation.

$y = 2 x^{2} + 3 x + 1$

Step 2

Interchange the variables.

$x = 2 y^{2} + 3 y + 1$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2 y^{2} + 3 y + 1 = x$ .

$2 y^{2} + 3 y + 1 = x$

Step 3.2

Subtract $x$ from both sides of the equation.

$2 y^{2} + 3 y + 1 - x = 0$

Step 3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.4

Substitute the values $a = 2$ , $b = 3$ , and $c = 1 - x$ into the quadratic formula and solve for $y$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (2 \cdot (1 - x))}}{2 \cdot 2}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

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

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot 2 \cdot (1 - x)}}{2 \cdot 2}$

Step 3.5.1.2

Multiply $- 4$ by $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 \cdot (1 - x)}}{2 \cdot 2}$

Step 3.5.1.3

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 8 \cdot 1 - 8 (- x)}}{2 \cdot 2}$

Step 3.5.1.4

Multiply $- 8$ by $1$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 - 8 (- x)}}{2 \cdot 2}$

Step 3.5.1.5

Multiply $- 1$ by $- 8$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 + 8 x}}{2 \cdot 2}$

Step 3.5.1.6

Subtract $8$ from $9$ .

$y = \frac{- 3 \pm \sqrt{1 + 8 x}}{2 \cdot 2}$

Step 3.5.2

Multiply $2$ by $2$ .

$y = \frac{- 3 \pm \sqrt{1 + 8 x}}{4}$

Step 3.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

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

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot 2 \cdot (1 - x)}}{2 \cdot 2}$

Step 3.6.1.2

Multiply $- 4$ by $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 \cdot (1 - x)}}{2 \cdot 2}$

Step 3.6.1.3

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 8 \cdot 1 - 8 (- x)}}{2 \cdot 2}$

Step 3.6.1.4

Multiply $- 8$ by $1$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 - 8 (- x)}}{2 \cdot 2}$

Step 3.6.1.5

Multiply $- 1$ by $- 8$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 + 8 x}}{2 \cdot 2}$

Step 3.6.1.6

Subtract $8$ from $9$ .

$y = \frac{- 3 \pm \sqrt{1 + 8 x}}{2 \cdot 2}$

Step 3.6.2

Multiply $2$ by $2$ .

$y = \frac{- 3 \pm \sqrt{1 + 8 x}}{4}$

Step 3.6.3

Change the $\pm$ to $+$ .

$y = \frac{- 3 + \sqrt{1 + 8 x}}{4}$

Step 3.6.4

Rewrite $- 3$ as $- 1 (3)$ .

$y = \frac{- 1 \cdot 3 + \sqrt{1 + 8 x}}{4}$

Step 3.6.5

Factor $- 1$ out of $\sqrt{1 + 8 x}$ .

$y = \frac{- 1 \cdot 3 - 1 (- \sqrt{1 + 8 x})}{4}$

Step 3.6.6

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{1 + 8 x})$ .

$y = \frac{- 1 (3 - \sqrt{1 + 8 x})}{4}$

Step 3.6.7

Move the negative in front of the fraction.

$y = - \frac{3 - \sqrt{1 + 8 x}}{4}$

Step 3.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.7.1

Simplify the numerator.

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Step 3.7.1.1

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

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot 2 \cdot (1 - x)}}{2 \cdot 2}$

Step 3.7.1.2

Multiply $- 4$ by $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 \cdot (1 - x)}}{2 \cdot 2}$

Step 3.7.1.3

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 8 \cdot 1 - 8 (- x)}}{2 \cdot 2}$

Step 3.7.1.4

Multiply $- 8$ by $1$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 - 8 (- x)}}{2 \cdot 2}$

Step 3.7.1.5

Multiply $- 1$ by $- 8$ .

$y = \frac{- 3 \pm \sqrt{9 - 8 + 8 x}}{2 \cdot 2}$

Step 3.7.1.6

Subtract $8$ from $9$ .

$y = \frac{- 3 \pm \sqrt{1 + 8 x}}{2 \cdot 2}$

Step 3.7.2

Multiply $2$ by $2$ .

$y = \frac{- 3 \pm \sqrt{1 + 8 x}}{4}$

Step 3.7.3

Change the $\pm$ to $-$ .

$y = \frac{- 3 - \sqrt{1 + 8 x}}{4}$

Step 3.7.4

Factor $- 1$ out of $- 3 - \sqrt{1 + 8 x}$ .

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Step 3.7.4.1

Reorder $1$ and $8 x$ .

$y = \frac{- 3 - \sqrt{8 x + 1}}{4}$

Step 3.7.4.2

Rewrite $- 3$ as $- 1 (3)$ .

$y = \frac{- 1 \cdot 3 - \sqrt{8 x + 1}}{4}$

Step 3.7.4.3

Factor $- 1$ out of $- \sqrt{8 x + 1}$ .

$y = \frac{- 1 \cdot 3 - (\sqrt{8 x + 1})}{4}$

Step 3.7.4.4

Factor $- 1$ out of $- 1 (3) - (\sqrt{8 x + 1})$ .

$y = \frac{- 1 (3 + \sqrt{8 x + 1})}{4}$

Step 3.7.4.5

Rewrite $- 1 (3 + \sqrt{8 x + 1})$ as $- (3 + \sqrt{8 x + 1})$ .

$y = \frac{- (3 + \sqrt{8 x + 1})}{4}$

Step 3.7.5

Move the negative in front of the fraction.

$y = - \frac{3 + \sqrt{8 x + 1}}{4}$

Step 3.8

The final answer is the combination of both solutions.

$y = - \frac{3 - \sqrt{1 + 8 x}}{4}$

$y = - \frac{3 + \sqrt{8 x + 1}}{4}$

$y = - \frac{3 - \sqrt{1 + 8 x}}{4}$

$y = - \frac{3 + \sqrt{8 x + 1}}{4}$

Step 4

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = - \frac{3 - \sqrt{1 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 1}}{4}$

Step 5

Verify if $f^{- 1} (x) = - \frac{3 - \sqrt{1 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 1}}{4}$ is the inverse of $f (x) = 2 x^{2} + 3 x + 1$ .

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Step 5.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 2 x^{2} + 3 x + 1$ and $f^{- 1} (x) = - \frac{3 - \sqrt{1 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 1}}{4}$ and compare them.

Step 5.2

Find the range of $f (x) = 2 x^{2} + 3 x + 1$ .

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Step 5.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[- \frac{1}{8}, \infty)$

Step 5.3

Find the domain of $- \frac{3 - \sqrt{1 + 8 x}}{4}$ .

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Step 5.3.1

Set the radicand in $\sqrt{1 + 8 x}$ greater than or equal to $0$ to find where the expression is defined.

$1 + 8 x \geq 0$

Step 5.3.2

Solve for $x$ .

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Step 5.3.2.1

Subtract $1$ from both sides of the inequality.

$8 x \geq - 1$

Step 5.3.2.2

Divide each term in $8 x \geq - 1$ by $8$ and simplify.

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Step 5.3.2.2.1

Divide each term in $8 x \geq - 1$ by $8$ .

$\frac{8 x}{8} \geq \frac{- 1}{8}$

Step 5.3.2.2.2

Simplify the left side.

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Step 5.3.2.2.2.1

Cancel the common factor of $8$ .

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Step 5.3.2.2.2.1.1

Cancel the common factor.

$\frac{8 x}{8} \geq \frac{- 1}{8}$

Step 5.3.2.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{- 1}{8}$

Step 5.3.2.2.3

Simplify the right side.

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Step 5.3.2.2.3.1

Move the negative in front of the fraction.

$x \geq - \frac{1}{8}$

Step 5.3.3

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

$[- \frac{1}{8}, \infty)$

Step 5.4

Find the domain of $f (x) = 2 x^{2} + 3 x + 1$ .

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Step 5.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Step 5.5

Since the domain of $f^{- 1} (x) = - \frac{3 - \sqrt{1 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 1}}{4}$ is the range of $f (x) = 2 x^{2} + 3 x + 1$ and the range of $f^{- 1} (x) = - \frac{3 - \sqrt{1 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 1}}{4}$ is the domain of $f (x) = 2 x^{2} + 3 x + 1$ , then $f^{- 1} (x) = - \frac{3 - \sqrt{1 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 1}}{4}$ is the inverse of $f (x) = 2 x^{2} + 3 x + 1$ .

$f^{- 1} (x) = - \frac{3 - \sqrt{1 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 1}}{4}$

Step 6