Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 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 $f (x) = x^{2} - 2 x + 1$ and $g (x) = x + 1$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $x^{2} - 2 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$ .

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

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 1.2.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

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

Step 1.2.4

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

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

Step 1.2.5

Multiply $- 2$ by $1$ .

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

Step 1.2.6

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

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

Step 1.2.7

Add $2 x - 2$ and $0$ .

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

Step 1.2.8

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

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

Step 1.2.9

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

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

Step 1.2.10

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

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

Step 1.2.11

Simplify the expression.

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Step 1.2.11.1

Add $1$ and $0$ .

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

Step 1.2.11.2

Multiply $- 1$ by $1$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

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

Step 1.3.2

Simplify the numerator.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Expand $(x + 1) (2 x - 2)$ using the FOIL Method.

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Step 1.3.2.1.1.1

Apply the distributive property.

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

Step 1.3.2.1.1.2

Apply the distributive property.

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

Step 1.3.2.1.1.3

Apply the distributive property.

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

Step 1.3.2.1.2

Simplify and combine like terms.

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Step 1.3.2.1.2.1

Simplify each term.

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Step 1.3.2.1.2.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.3.2.1.2.1.2

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

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Step 1.3.2.1.2.1.2.1

Move $x$ .

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

Step 1.3.2.1.2.1.2.2

Multiply $x$ by $x$ .

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

Step 1.3.2.1.2.1.3

Move $- 2$ to the left of $x$ .

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

Step 1.3.2.1.2.1.4

Multiply $2 x$ by $1$ .

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

Step 1.3.2.1.2.1.5

Multiply $- 2$ by $1$ .

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

Step 1.3.2.1.2.2

Add $- 2 x$ and $2 x$ .

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

Step 1.3.2.1.2.3

Add $2 x^{2}$ and $0$ .

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

Step 1.3.2.1.3

Multiply $- 2$ by $- 1$ .

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

Step 1.3.2.1.4

Multiply $- 1$ by $1$ .

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

Step 1.3.2.2

Subtract $x^{2}$ from $2 x^{2}$ .

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

Step 1.3.2.3

Subtract $1$ from $- 2$ .

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

Step 1.3.3

Factor $x^{2} + 2 x - 3$ using the AC method.

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Step 1.3.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Step 1.3.3.2

Write the factored form using these integers.

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

Step 2

Find the second derivative of the function.

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Step 2.1

$f'' (x) = \frac{{(x + 1)}^{2} \frac{d}{d x} ((x - 1) (x + 3)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{({(x + 1)}^{2})}^{2}}$

Step 2.2

Multiply the exponents in ${({(x + 1)}^{2})}^{2}$ .

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Step 2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = \frac{{(x + 1)}^{2} \frac{d}{d x} ((x - 1) (x + 3)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{2 \cdot 2}}$

Step 2.2.2

Multiply $2$ by $2$ .

$f'' (x) = \frac{{(x + 1)}^{2} \frac{d}{d x} ((x - 1) (x + 3)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x - 1$ and $g (x) = x + 3$ .

$f'' (x) = \frac{{(x + 1)}^{2} ((x - 1) \frac{d}{d x} (x + 3) + (x + 3) \frac{d}{d x} (x - 1)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4

Differentiate.

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Step 2.4.1

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

$f'' (x) = \frac{{(x + 1)}^{2} ((x - 1) (\frac{d}{d x} (x) + \frac{d}{d x} (3)) + (x + 3) \frac{d}{d x} (x - 1)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.2

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

$f'' (x) = \frac{{(x + 1)}^{2} ((x - 1) (1 + \frac{d}{d x} (3)) + (x + 3) \frac{d}{d x} (x - 1)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.3

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

$f'' (x) = \frac{{(x + 1)}^{2} ((x - 1) (1 + 0) + (x + 3) \frac{d}{d x} (x - 1)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.4

Simplify the expression.

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Step 2.4.4.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x + 1)}^{2} ((x - 1) \cdot 1 + (x + 3) \frac{d}{d x} (x - 1)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.4.2

Multiply $x - 1$ by $1$ .

$f'' (x) = \frac{{(x + 1)}^{2} (x - 1 + (x + 3) \frac{d}{d x} (x - 1)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.5

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

$f'' (x) = \frac{{(x + 1)}^{2} (x - 1 + (x + 3) (\frac{d}{d x} (x) + \frac{d}{d x} (- 1))) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.6

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

$f'' (x) = \frac{{(x + 1)}^{2} (x - 1 + (x + 3) (1 + \frac{d}{d x} (- 1))) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.7

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

$f'' (x) = \frac{{(x + 1)}^{2} (x - 1 + (x + 3) (1 + 0)) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.8

Simplify by adding terms.

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Step 2.4.8.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x + 1)}^{2} (x - 1 + (x + 3) \cdot 1) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.8.2

Multiply $x + 3$ by $1$ .

$f'' (x) = \frac{{(x + 1)}^{2} (x - 1 + x + 3) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.8.3

Add $x$ and $x$ .

$f'' (x) = \frac{{(x + 1)}^{2} (2 x - 1 + 3) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.4.8.4

Add $- 1$ and $3$ .

$f'' (x) = \frac{{(x + 1)}^{2} (2 x + 2) - (x - 1) (x + 3) \frac{d}{d x} {(x + 1)}^{2}}{{(x + 1)}^{4}}$

Step 2.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = x + 1$ .

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Step 2.5.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$f'' (x) = \frac{{(x + 1)}^{2} (2 x + 2) - (x - 1) (x + 3) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x + 1))}{{(x + 1)}^{4}}$

Step 2.5.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$f'' (x) = \frac{{(x + 1)}^{2} (2 x + 2) - (x - 1) (x + 3) (2 u \frac{d}{d x} (x + 1))}{{(x + 1)}^{4}}$

Step 2.5.3

Replace all occurrences of $u$ with $x + 1$ .

$f'' (x) = \frac{{(x + 1)}^{2} (2 x + 2) - (x - 1) (x + 3) (2 (x + 1) \frac{d}{d x} (x + 1))}{{(x + 1)}^{4}}$

Step 2.6

Simplify with factoring out.

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Step 2.6.1

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x + 1)}^{2} (2 x + 2) - 2 (x - 1) (x + 3) ((x + 1) \frac{d}{d x} (x + 1))}{{(x + 1)}^{4}}$

Step 2.6.2

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

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Step 2.6.2.1

Factor $x + 1$ out of ${(x + 1)}^{2} (2 x + 2)$ .

$f'' (x) = \frac{(x + 1) ((x + 1) (2 x + 2)) - 2 (x - 1) (x + 3) ((x + 1) \frac{d}{d x} (x + 1))}{{(x + 1)}^{4}}$

Step 2.6.2.2

Factor $x + 1$ out of $- 2 (x - 1) (x + 3) ((x + 1) \frac{d}{d x} [x + 1])$ .

$f'' (x) = \frac{(x + 1) ((x + 1) (2 x + 2)) + (x + 1) (- 2 (x - 1) (x + 3) (\frac{d}{d x} (x + 1)))}{{(x + 1)}^{4}}$

Step 2.6.2.3

Factor $x + 1$ out of $(x + 1) ((x + 1) (2 x + 2)) + (x + 1) (- 2 (x - 1) (x + 3) (\frac{d}{d x} [x + 1]))$ .

$f'' (x) = \frac{(x + 1) ((x + 1) (2 x + 2) - 2 (x - 1) (x + 3) (\frac{d}{d x} (x + 1)))}{{(x + 1)}^{4}}$

Step 2.7

Cancel the common factors.

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Step 2.7.1

Factor $x + 1$ out of ${(x + 1)}^{4}$ .

$f'' (x) = \frac{(x + 1) ((x + 1) (2 x + 2) - 2 (x - 1) (x + 3) (\frac{d}{d x} (x + 1)))}{(x + 1) {(x + 1)}^{3}}$

Step 2.7.2

Cancel the common factor.

$f'' (x) = \frac{(x + 1) ((x + 1) (2 x + 2) - 2 (x - 1) (x + 3) (\frac{d}{d x} (x + 1)))}{(x + 1) {(x + 1)}^{3}}$

Step 2.7.3

Rewrite the expression.

$f'' (x) = \frac{(x + 1) (2 x + 2) - 2 (x - 1) (x + 3) (\frac{d}{d x} (x + 1))}{{(x + 1)}^{3}}$

Step 2.8

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

$f'' (x) = \frac{(x + 1) (2 x + 2) - 2 (x - 1) (x + 3) (\frac{d}{d x} (x) + \frac{d}{d x} (1))}{{(x + 1)}^{3}}$

Step 2.9

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

$f'' (x) = \frac{(x + 1) (2 x + 2) - 2 (x - 1) (x + 3) (1 + \frac{d}{d x} (1))}{{(x + 1)}^{3}}$

Step 2.10

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

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

Step 2.11

Simplify the expression.

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Step 2.11.1

Add $1$ and $0$ .

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

Step 2.11.2

Multiply $- 2$ by $1$ .

$f'' (x) = \frac{(x + 1) (2 x + 2) - 2 (x - 1) (x + 3)}{{(x + 1)}^{3}}$

Step 2.12

Simplify.

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Step 2.12.1

Apply the distributive property.

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

Step 2.12.2

Simplify the numerator.

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Step 2.12.2.1

Simplify each term.

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Step 2.12.2.1.1

Expand $(x + 1) (2 x + 2)$ using the FOIL Method.

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Step 2.12.2.1.1.1

Apply the distributive property.

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

Step 2.12.2.1.1.2

Apply the distributive property.

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

Step 2.12.2.1.1.3

Apply the distributive property.

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

Step 2.12.2.1.2

Simplify and combine like terms.

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Step 2.12.2.1.2.1

Simplify each term.

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Step 2.12.2.1.2.1.1

Rewrite using the commutative property of multiplication.

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

Step 2.12.2.1.2.1.2

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

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Step 2.12.2.1.2.1.2.1

Move $x$ .

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

Step 2.12.2.1.2.1.2.2

Multiply $x$ by $x$ .

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

Step 2.12.2.1.2.1.3

Move $2$ to the left of $x$ .

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

Step 2.12.2.1.2.1.4

Multiply $2 x$ by $1$ .

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

Step 2.12.2.1.2.1.5

Multiply $2$ by $1$ .

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

Step 2.12.2.1.2.2

Add $2 x$ and $2 x$ .

$f'' (x) = \frac{2 x^{2} + 4 x + 2 + (- 2 x - 2 \cdot - 1) (x + 3)}{{(x + 1)}^{3}}$

Step 2.12.2.1.3

Multiply $- 2$ by $- 1$ .

$f'' (x) = \frac{2 x^{2} + 4 x + 2 + (- 2 x + 2) (x + 3)}{{(x + 1)}^{3}}$

Step 2.12.2.1.4

Expand $(- 2 x + 2) (x + 3)$ using the FOIL Method.

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Step 2.12.2.1.4.1

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 x (x + 3) + 2 (x + 3)}{{(x + 1)}^{3}}$

Step 2.12.2.1.4.2

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 x \cdot x - 2 x \cdot 3 + 2 (x + 3)}{{(x + 1)}^{3}}$

Step 2.12.2.1.4.3

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 x \cdot x - 2 x \cdot 3 + 2 x + 2 \cdot 3}{{(x + 1)}^{3}}$

Step 2.12.2.1.5

Simplify and combine like terms.

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Step 2.12.2.1.5.1

Simplify each term.

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Step 2.12.2.1.5.1.1

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

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Step 2.12.2.1.5.1.1.1

Move $x$ .

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 (x \cdot x) - 2 x \cdot 3 + 2 x + 2 \cdot 3}{{(x + 1)}^{3}}$

Step 2.12.2.1.5.1.1.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 x^{2} - 2 x \cdot 3 + 2 x + 2 \cdot 3}{{(x + 1)}^{3}}$

Step 2.12.2.1.5.1.2

Multiply $3$ by $- 2$ .

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 x^{2} - 6 x + 2 x + 2 \cdot 3}{{(x + 1)}^{3}}$

Step 2.12.2.1.5.1.3

Multiply $2$ by $3$ .

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 x^{2} - 6 x + 2 x + 6}{{(x + 1)}^{3}}$

Step 2.12.2.1.5.2

Add $- 6 x$ and $2 x$ .

$f'' (x) = \frac{2 x^{2} + 4 x + 2 - 2 x^{2} - 4 x + 6}{{(x + 1)}^{3}}$

Step 2.12.2.2

Combine the opposite terms in $2 x^{2} + 4 x + 2 - 2 x^{2} - 4 x + 6$ .

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Step 2.12.2.2.1

Subtract $2 x^{2}$ from $2 x^{2}$ .

$f'' (x) = \frac{4 x + 2 + 0 - 4 x + 6}{{(x + 1)}^{3}}$

Step 2.12.2.2.2

Add $4 x + 2$ and $0$ .

$f'' (x) = \frac{4 x + 2 - 4 x + 6}{{(x + 1)}^{3}}$

Step 2.12.2.2.3

Subtract $4 x$ from $4 x$ .

$f'' (x) = \frac{0 + 2 + 6}{{(x + 1)}^{3}}$

Step 2.12.2.2.4

Add $0$ and $2$ .

$f'' (x) = \frac{2 + 6}{{(x + 1)}^{3}}$

Step 2.12.2.3

Add $2$ and $6$ .

$f'' (x) = \frac{8}{{(x + 1)}^{3}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

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

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $x^{2} - 2 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$ .

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

Step 4.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 4.1.2.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

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

Step 4.1.2.4

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

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

Step 4.1.2.5

Multiply $- 2$ by $1$ .

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

Step 4.1.2.6

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

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

Step 4.1.2.7

Add $2 x - 2$ and $0$ .

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

Step 4.1.2.8

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

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

Step 4.1.2.9

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

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

Step 4.1.2.10

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

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

Step 4.1.2.11

Simplify the expression.

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Step 4.1.2.11.1

Add $1$ and $0$ .

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

Step 4.1.2.11.2

Multiply $- 1$ by $1$ .

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

Step 4.1.3

Simplify.

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Step 4.1.3.1

Apply the distributive property.

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

Step 4.1.3.2

Simplify the numerator.

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Step 4.1.3.2.1

Simplify each term.

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Step 4.1.3.2.1.1

Expand $(x + 1) (2 x - 2)$ using the FOIL Method.

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Step 4.1.3.2.1.1.1

Apply the distributive property.

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

Step 4.1.3.2.1.1.2

Apply the distributive property.

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

Step 4.1.3.2.1.1.3

Apply the distributive property.

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

Step 4.1.3.2.1.2

Simplify and combine like terms.

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Step 4.1.3.2.1.2.1

Simplify each term.

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Step 4.1.3.2.1.2.1.1

Rewrite using the commutative property of multiplication.

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

Step 4.1.3.2.1.2.1.2

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

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Step 4.1.3.2.1.2.1.2.1

Move $x$ .

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

Step 4.1.3.2.1.2.1.2.2

Multiply $x$ by $x$ .

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

Step 4.1.3.2.1.2.1.3

Move $- 2$ to the left of $x$ .

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

Step 4.1.3.2.1.2.1.4

Multiply $2 x$ by $1$ .

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

Step 4.1.3.2.1.2.1.5

Multiply $- 2$ by $1$ .

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

Step 4.1.3.2.1.2.2

Add $- 2 x$ and $2 x$ .

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

Step 4.1.3.2.1.2.3

Add $2 x^{2}$ and $0$ .

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

Step 4.1.3.2.1.3

Multiply $- 2$ by $- 1$ .

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

Step 4.1.3.2.1.4

Multiply $- 1$ by $1$ .

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

Step 4.1.3.2.2

Subtract $x^{2}$ from $2 x^{2}$ .

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

Step 4.1.3.2.3

Subtract $1$ from $- 2$ .

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

Step 4.1.3.3

Factor $x^{2} + 2 x - 3$ using the AC method.

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Step 4.1.3.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Step 4.1.3.3.2

Write the factored form using these integers.

$f' (x) = \frac{(x - 1) (x + 3)}{{(x + 1)}^{2}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{(x - 1) (x + 3)}{{(x + 1)}^{2}}$ .

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

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{(x - 1) (x + 3)}{{(x + 1)}^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

Set the numerator equal to zero.

$(x - 1) (x + 3) = 0$

Step 5.3

Solve the equation for $x$ .

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Step 5.3.1

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

$x - 1 = 0$

$x + 3 = 0$

Step 5.3.2

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

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Step 5.3.2.1

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

$x - 1 = 0$

Step 5.3.2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5.3.3

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 5.3.3.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 5.3.3.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5.3.4

The final solution is all the values that make $(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{(x - 1) (x + 3)}{{(x + 1)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Step 6.2

Solve for $x$ .

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Step 6.2.1

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

$x + 1 = 0$

Step 6.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

Critical points to evaluate.

$x = 1, - 3$

Step 8

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{8}{{((1) + 1)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the denominator.

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Step 9.1.1

Add $1$ and $1$ .

$\frac{8}{2^{3}}$

Step 9.1.2

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

$\frac{8}{8}$

Step 9.2

Divide $8$ by $8$ .

$1$

Step 10

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Step 11

Find the y-value when $x = 1$ .

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Step 11.1

Replace the variable $x$ with $1$ in the expression.

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

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify the numerator.

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Step 11.2.1.1

One to any power is one.

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

Step 11.2.1.2

Multiply $- 2$ by $1$ .

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

Step 11.2.1.3

Subtract $2$ from $1$ .

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

Step 11.2.1.4

Add $- 1$ and $1$ .

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

Step 11.2.2

Simplify the expression.

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Step 11.2.2.1

Add $1$ and $1$ .

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

Step 11.2.2.2

Divide $0$ by $2$ .

$f (1) = 0$

Step 11.2.3

The final answer is $0$ .

$y = 0$

Step 12

Evaluate the second derivative at $x = - 3$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{8}{{((- 3) + 1)}^{3}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify the denominator.

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Step 13.1.1

Add $- 3$ and $1$ .

$\frac{8}{{(- 2)}^{3}}$

Step 13.1.2

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

$\frac{8}{- 8}$

Step 13.2

Divide $8$ by $- 8$ .

$- 1$

Step 14

$x = - 3$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 3$ is a local maximum

Step 15

Find the y-value when $x = - 3$ .

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Step 15.1

Replace the variable $x$ with $- 3$ in the expression.

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

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify the numerator.

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Step 15.2.1.1

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

$f (- 3) = \frac{9 - 2 \cdot - 3 + 1}{- 3 + 1}$

Step 15.2.1.2

Multiply $- 2$ by $- 3$ .

$f (- 3) = \frac{9 + 6 + 1}{- 3 + 1}$

Step 15.2.1.3

Add $9$ and $6$ .

$f (- 3) = \frac{15 + 1}{- 3 + 1}$

Step 15.2.1.4

Add $15$ and $1$ .

$f (- 3) = \frac{16}{- 3 + 1}$

Step 15.2.2

Simplify the expression.

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Step 15.2.2.1

Add $- 3$ and $1$ .

$f (- 3) = \frac{16}{- 2}$

Step 15.2.2.2

Divide $16$ by $- 2$ .

$f (- 3) = - 8$

Step 15.2.3

The final answer is $- 8$ .

$y = - 8$

Step 16

These are the local extrema for $f (x) = \frac{x^{2} - 2 x + 1}{x + 1}$ .

$(1, 0)$ is a local minima

$(- 3, - 8)$ is a local maxima

Step 17