Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = 1 - x$ .

$\frac{(1 - x) \frac{d}{d x} [x] - x \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{(1 - x) \cdot 1 - x \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.1.2.2

Multiply $1 - x$ by $1$ .

$\frac{1 - x - x \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.1.2.3

By the Sum Rule, the derivative of $1 - x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x]$ .

$\frac{1 - x - x (\frac{d}{d x} [1] + \frac{d}{d x} [- x])}{{(1 - x)}^{2}}$

Step 1.1.2.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{1 - x - x (0 + \frac{d}{d x} [- x])}{{(1 - x)}^{2}}$

Step 1.1.2.5

Add $0$ and $\frac{d}{d x} [- x]$ .

$\frac{1 - x - x \frac{d}{d x} [- x]}{{(1 - x)}^{2}}$

Step 1.1.2.6

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$\frac{1 - x - x (- \frac{d}{d x} [x])}{{(1 - x)}^{2}}$

Step 1.1.2.7

Multiply.

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Step 1.1.2.7.1

Multiply $- 1$ by $- 1$ .

$\frac{1 - x + 1 x \frac{d}{d x} [x]}{{(1 - x)}^{2}}$

Step 1.1.2.7.2

Multiply $x$ by $1$ .

$\frac{1 - x + x \frac{d}{d x} [x]}{{(1 - x)}^{2}}$

Step 1.1.2.8

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

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

Step 1.1.2.9

Simplify by adding terms.

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Step 1.1.2.9.1

Multiply $x$ by $1$ .

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

Step 1.1.2.9.2

Add $- x$ and $x$ .

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

Step 1.1.2.9.3

Simplify the expression.

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Step 1.1.2.9.3.1

Add $1$ and $0$ .

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

Step 1.1.2.9.3.2

Reorder terms.

$f' (x) = \frac{1}{{(- x + 1)}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{1}{{(- x + 1)}^{2}}$ .

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

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{1}{{(- x + 1)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

$1 = 0$

Step 2.3

Since $1 \neq 0$ , there are no solutions.

No solution

Step 3

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{1}{{(- x + 1)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Step 4.2

Solve for $x$ .

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Step 4.2.1

Factor the left side of the equation.

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Step 4.2.1.1

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

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Step 4.2.1.1.1

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

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

Step 4.2.1.1.2

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

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

Step 4.2.1.1.3

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

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

Step 4.2.1.2

Apply the product rule to $- (x - 1)$ .

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

Step 4.2.2

Divide each term in ${(- 1)}^{2} {(x - 1)}^{2} = 0$ by ${(- 1)}^{2}$ and simplify.

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Step 4.2.2.1

Divide each term in ${(- 1)}^{2} {(x - 1)}^{2} = 0$ by ${(- 1)}^{2}$ .

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of ${(- 1)}^{2}$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

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

Step 4.2.2.2.1.2

Divide ${(x - 1)}^{2}$ by $1$ .

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

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

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

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

Step 4.2.2.3.2

Divide $0$ by $1$ .

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

Step 4.2.3

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

$x - 1 = 0$

Step 4.2.4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

After finding the point that makes the derivative $f' (x) = \frac{1}{{(- x + 1)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{x}{1 - x}$ is increasing and where it is decreasing is $(- \infty, 1) \cup (1, \infty)$ .

$(- \infty, 1) \cup (1, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0$ in the expression.

$f' (0) = \frac{1}{{(- (0) + 1)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the denominator.

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Step 6.2.1.1

Multiply $- 1$ by $0$ .

$f' (0) = \frac{1}{{(0 + 1)}^{2}}$

Step 6.2.1.2

Add $0$ and $1$ .

$f' (0) = \frac{1}{1^{2}}$

Step 6.2.1.3

One to any power is one.

$f' (0) = \frac{1}{1}$

Step 6.2.2

Divide $1$ by $1$ .

$f' (0) = 1$

Step 6.2.3

The final answer is $1$ .

$1$

Step 6.3

At $x = 0$ the derivative is $1$ . Since this is positive, the function is increasing on $(- \infty, 1)$ .

Increasing on $(- \infty, 1)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(1, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $2$ in the expression.

$f' (2) = \frac{1}{{(- (2) + 1)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the denominator.

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Step 7.2.1.1

Multiply $- 1$ by $2$ .

$f' (2) = \frac{1}{{(- 2 + 1)}^{2}}$

Step 7.2.1.2

Add $- 2$ and $1$ .

$f' (2) = \frac{1}{{(- 1)}^{2}}$

Step 7.2.1.3

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

$f' (2) = \frac{1}{1}$

Step 7.2.2

Divide $1$ by $1$ .

$f' (2) = 1$

Step 7.2.3

The final answer is $1$ .

$1$

Step 7.3

At $x = 2$ the derivative is $1$ . Since this is positive, the function is increasing on $(1, \infty)$ .

Increasing on $(1, \infty)$ since $f' (x) > 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, 1), (1, \infty)$

Step 9