Calculus Examples

$f (x) = x^{4} + 5 x^{2}$ , $a = - 1$

Step 1

Consider the function used to find the linearization at $a$ .

$L (x) = f (a) + f' (a) (x - a)$

Step 2

Substitute the value of $a = - 1$ into the linearization function.

$L (x) = f (- 1) + f' (- 1) (x - - 1)$

Step 3

Evaluate $f (- 1)$ .

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Step 3.1

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

$f (- 1) = {(- 1)}^{4} + 5 {(- 1)}^{2}$

Step 3.2

Simplify ${(- 1)}^{4} + 5 {(- 1)}^{2}$ .

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Step 3.2.1

Remove parentheses.

${(- 1)}^{4} + 5 {(- 1)}^{2}$

Step 3.2.2

Simplify each term.

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Step 3.2.2.1

Raise $- 1$ to the power of $4$ .

$1 + 5 {(- 1)}^{2}$

Step 3.2.2.2

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

$1 + 5 \cdot 1$

Step 3.2.2.3

Multiply $5$ by $1$ .

$1 + 5$

Step 3.2.3

Add $1$ and $5$ .

$6$

Step 4

Find the derivative and evaluate it at $- 1$ .

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Step 4.1

Find the derivative of $f (x) = x^{4} + 5 x^{2}$ .

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Step 4.1.1

Differentiate.

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Step 4.1.1.1

By the Sum Rule, the derivative of $x^{4} + 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [5 x^{2}]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [5 x^{2}]$

Step 4.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$4 x^{3} + \frac{d}{d x} [5 x^{2}]$

Step 4.1.2

Evaluate $\frac{d}{d x} [5 x^{2}]$ .

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Step 4.1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} + 5 \frac{d}{d x} [x^{2}]$

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 $n = 2$ .

$4 x^{3} + 5 (2 x)$

Step 4.1.2.3

Multiply $2$ by $5$ .

$4 x^{3} + 10 x$

Step 4.2

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

$4 {(- 1)}^{3} + 10 (- 1)$

Step 4.3

Simplify.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Raise $- 1$ to the power of $3$ .

$4 \cdot - 1 + 10 (- 1)$

Step 4.3.1.2

Multiply $4$ by $- 1$ .

$- 4 + 10 (- 1)$

Step 4.3.1.3

Multiply $10$ by $- 1$ .

$- 4 - 10$

Step 4.3.2

Subtract $10$ from $- 4$ .

$- 14$

Step 5

Substitute the components into the linearization function in order to find the linearization at $a$ .

$L (x) = 6 - 14 (x - - 1)$

Step 6

Simplify.

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Step 6.1

Simplify each term.

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Step 6.1.1

Apply the distributive property.

$L (x) = 6 - 14 x - 14 \cdot 1$

Step 6.1.2

Multiply $- 14$ by $1$ .

$L (x) = 6 - 14 x - 14$

Step 6.2

Subtract $14$ from $6$ .

$L (x) = - 14 x - 8$

Step 7