Trigonometry Examples

$f (\frac{1}{x}) = x - 1$

Step 1

Simplify $f (\frac{1}{x})$ .

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Step 1.1

Rewrite.

$0 + 0 + f (\frac{1}{x}) = x - 1$

Step 1.2

Simplify by adding zeros.

$f (\frac{1}{x}) = x - 1$

Step 1.3

Combine $f$ and $\frac{1}{x}$ .

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

Step 2

Subtract $x$ from both sides of the equation.

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

Step 3

Find the LCD of the terms in the equation.

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Step 3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x, 1, 1$

Step 3.2

The LCM of one and any expression is the expression.

$x$

Step 4

Multiply each term in $\frac{f}{x} - x = - 1$ by $x$ to eliminate the fractions.

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Step 4.1

Multiply each term in $\frac{f}{x} - x = - 1$ by $x$ .

$\frac{f}{x} x - x \cdot x = - x$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Cancel the common factor of $x$ .

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Step 4.2.1.1.1

Cancel the common factor.

$\frac{f}{x} x - x \cdot x = - x$

Step 4.2.1.1.2

Rewrite the expression.

$f - x \cdot x = - x$

Step 4.2.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 4.2.1.2.1

Move $x$ .

$f - (x \cdot x) = - x$

Step 4.2.1.2.2

Multiply $x$ by $x$ .

$f - x^{2} = - x$

Step 5

Solve the equation.

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Step 5.1

Add $x$ to both sides of the equation.

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

Step 5.2

Use the quadratic formula to find the solutions.

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

Step 5.3

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

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

Step 5.4

Simplify.

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Step 5.4.1

Simplify the numerator.

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Step 5.4.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot f}}{2 \cdot - 1}$

Step 5.4.1.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 f}}{2 \cdot - 1}$

Step 5.4.2

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 f}}{- 2}$

Step 5.4.3

Simplify $\frac{- 1 \pm \sqrt{1 + 4 f}}{- 2}$ .

$x = \frac{1 \pm \sqrt{1 + 4 f}}{2}$

Step 5.5

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{1 + 4 f}}{2}$

$x = \frac{1 - \sqrt{1 + 4 f}}{2}$

$x = \frac{1 + \sqrt{1 + 4 f}}{2}$

$x = \frac{1 - \sqrt{1 + 4 f}}{2}$