Calculus Examples

$y = \frac{2 + x}{3 - x}$

Step 1

Write $y = \frac{2 + x}{3 - x}$ as a function.

$f (x) = \frac{2 + x}{3 - x}$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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 $f (x) = 2 + x$ and $g (x) = 3 - x$ .

$\frac{(3 - x) \frac{d}{d x} [2 + x] - (2 + x) \frac{d}{d x} [3 - x]}{{(3 - x)}^{2}}$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $2 + x$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [x]$ .

$\frac{(3 - x) (\frac{d}{d x} [2] + \frac{d}{d x} [x]) - (2 + x) \frac{d}{d x} [3 - x]}{{(3 - x)}^{2}}$

Step 2.1.2.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{(3 - x) (0 + \frac{d}{d x} [x]) - (2 + x) \frac{d}{d x} [3 - x]}{{(3 - x)}^{2}}$

Step 2.1.2.3

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

$\frac{(3 - x) \frac{d}{d x} [x] - (2 + x) \frac{d}{d x} [3 - x]}{{(3 - x)}^{2}}$

Step 2.1.2.4

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

$\frac{(3 - x) \cdot 1 - (2 + x) \frac{d}{d x} [3 - x]}{{(3 - x)}^{2}}$

Step 2.1.2.5

Multiply $3 - x$ by $1$ .

$\frac{3 - x - (2 + x) \frac{d}{d x} [3 - x]}{{(3 - x)}^{2}}$

Step 2.1.2.6

By the Sum Rule, the derivative of $3 - x$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [- x]$ .

$\frac{3 - x - (2 + x) (\frac{d}{d x} [3] + \frac{d}{d x} [- x])}{{(3 - x)}^{2}}$

Step 2.1.2.7

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$\frac{3 - x - (2 + x) (0 + \frac{d}{d x} [- x])}{{(3 - x)}^{2}}$

Step 2.1.2.8

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

$\frac{3 - x - (2 + x) \frac{d}{d x} [- x]}{{(3 - x)}^{2}}$

Step 2.1.2.9

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

$\frac{3 - x - (2 + x) (- \frac{d}{d x} [x])}{{(3 - x)}^{2}}$

Step 2.1.2.10

Multiply.

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Step 2.1.2.10.1

Multiply $- 1$ by $- 1$ .

$\frac{3 - x + 1 (2 + x) \frac{d}{d x} [x]}{{(3 - x)}^{2}}$

Step 2.1.2.10.2

Multiply $2 + x$ by $1$ .

$\frac{3 - x + (2 + x) \frac{d}{d x} [x]}{{(3 - x)}^{2}}$

Step 2.1.2.11

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

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

Step 2.1.2.12

Simplify by adding terms.

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Step 2.1.2.12.1

Multiply $2 + x$ by $1$ .

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

Step 2.1.2.12.2

Add $3$ and $2$ .

$\frac{- x + 5 + x}{{(3 - x)}^{2}}$

Step 2.1.2.12.3

Add $- x$ and $x$ .

$\frac{0 + 5}{{(3 - x)}^{2}}$

Step 2.1.2.12.4

Simplify the expression.

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Step 2.1.2.12.4.1

Add $0$ and $5$ .

$\frac{5}{{(3 - x)}^{2}}$

Step 2.1.2.12.4.2

Reorder terms.

$f' (x) = \frac{5}{{(- x + 3)}^{2}}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{5}{{(- x + 3)}^{2}}$ .

$\frac{5}{{(- x + 3)}^{2}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{5}{{(- x + 3)}^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{5}{{(- x + 3)}^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$5 = 0$

Step 3.3

Since $5 \neq 0$ , there are no solutions.

No solution

Step 4

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 5

Find where the derivative is undefined.

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Step 5.1

Set the denominator in $\frac{5}{{(- x + 3)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(- x + 3)}^{2} = 0$

Step 5.2

Solve for $x$ .

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Step 5.2.1

Factor the left side of the equation.

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Step 5.2.1.1

Factor $- 1$ out of $- x + 3$ .

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Step 5.2.1.1.1

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

${(- (x) + 3)}^{2} = 0$

Step 5.2.1.1.2

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

${(- (x) - 1 \cdot - 3)}^{2} = 0$

Step 5.2.1.1.3

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

${(- (x - 3))}^{2} = 0$

Step 5.2.1.2

Apply the product rule to $- (x - 3)$ .

${(- 1)}^{2} {(x - 3)}^{2} = 0$

Step 5.2.2

Divide each term in ${(- 1)}^{2} {(x - 3)}^{2} = 0$ by ${(- 1)}^{2}$ and simplify.

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Step 5.2.2.1

Divide each term in ${(- 1)}^{2} {(x - 3)}^{2} = 0$ by ${(- 1)}^{2}$ .

$\frac{{(- 1)}^{2} {(x - 3)}^{2}}{{(- 1)}^{2}} = \frac{0}{{(- 1)}^{2}}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

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

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{{(- 1)}^{2} {(x - 3)}^{2}}{{(- 1)}^{2}} = \frac{0}{{(- 1)}^{2}}$

Step 5.2.2.2.1.2

Divide ${(x - 3)}^{2}$ by $1$ .

${(x - 3)}^{2} = \frac{0}{{(- 1)}^{2}}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

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

${(x - 3)}^{2} = \frac{0}{1}$

Step 5.2.2.3.2

Divide $0$ by $1$ .

${(x - 3)}^{2} = 0$

Step 5.2.3

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 5.2.4

Add $3$ to both sides of the equation.

$x = 3$

Step 6

After finding the point that makes the derivative $f' (x) = \frac{5}{{(- x + 3)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{2 + x}{3 - x}$ is increasing and where it is decreasing is $(- \infty, 3) \cup (3, \infty)$ .

$(- \infty, 3) \cup (3, \infty)$

Step 7

Substitute a value from the interval $(- \infty, 3)$ 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.

$f' (2) = \frac{5}{{(- (2) + 3)}^{2}}$

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 $2$ .

$f' (2) = \frac{5}{{(- 2 + 3)}^{2}}$

Step 7.2.1.2

Add $- 2$ and $3$ .

$f' (2) = \frac{5}{1^{2}}$

Step 7.2.1.3

One to any power is one.

$f' (2) = \frac{5}{1}$

Step 7.2.2

Divide $5$ by $1$ .

$f' (2) = 5$

Step 7.2.3

The final answer is $5$ .

$5$

Step 7.3

At $x = 2$ the derivative is $5$ . Since this is positive, the function is increasing on $(- \infty, 3)$ .

Increasing on $(- \infty, 3)$ since $f' (x) > 0$

Step 8

Substitute a value from the interval $(3, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = \frac{5}{{(- (4) + 3)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify the denominator.

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Step 8.2.1.1

Multiply $- 1$ by $4$ .

$f' (4) = \frac{5}{{(- 4 + 3)}^{2}}$

Step 8.2.1.2

Add $- 4$ and $3$ .

$f' (4) = \frac{5}{{(- 1)}^{2}}$

Step 8.2.1.3

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

$f' (4) = \frac{5}{1}$

Step 8.2.2

Divide $5$ by $1$ .

$f' (4) = 5$

Step 8.2.3

The final answer is $5$ .

$5$

Step 8.3

At $x = 4$ the derivative is $5$ . Since this is positive, the function is increasing on $(3, \infty)$ .

Increasing on $(3, \infty)$ since $f' (x) > 0$

Step 9

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

Increasing on: $(- \infty, 3), (3, \infty)$

Step 10