Calculus Examples

$t (x) = x^{3} - 5 x^{2} - 9 x + 45$ , $x - 5$

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 $a = x^{3} - 5 x^{2} - 9 x + 45$ into the linearization function.

$L (x) = f (x^{3} - 5 x^{2} - 9 x + 45) + f' (x^{3} - 5 x^{2} - 9 x + 45) (x - x^{3} - 5 x^{2} - 9 x + 45)$

Step 3

Evaluate $f (x^{3} - 5 x^{2} - 9 x + 45)$ .

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Step 3.1

Replace the variable $x$ with $x^{3} - 5 x^{2} - 9 x + 45$ in the expression.

$f (x^{3} - 5 x^{2} - 9 x + 45) = (x^{3} - 5 x^{2} - 9 x + 45) - 5$

Step 3.2

Simplify $(x^{3} - 5 x^{2} - 9 x + 45) - 5$ .

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Step 3.2.1

Remove parentheses.

$(x^{3} - 5 x^{2} - 9 x + 45) - 5$

Step 3.2.2

Subtract $5$ from $45$ .

$x^{3} - 5 x^{2} - 9 x + 40$

Step 4

Find the derivative of $f (x) = x - 5$ .

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Step 4.1

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

$\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$

Step 4.2

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

$1 + \frac{d}{d x} [- 5]$

Step 4.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$1 + 0$

Step 4.4

Add $1$ and $0$ .

$1$

Step 5

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

$L (x) = x^{3} - 5 x^{2} - 9 x + 40 + 1 (x - x^{3} - 5 x^{2} - 9 x + 45)$

Step 6

Simplify.

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Step 6.1

Simplify each term.

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Step 6.1.1

Multiply $x - x^{3} - 5 x^{2} - 9 x + 45$ by $1$ .

$L (x) = x^{3} - 5 x^{2} - 9 x + 40 + x - x^{3} - 5 x^{2} - 9 x + 45$

Step 6.1.2

Subtract $9 x$ from $x$ .

$L (x) = x^{3} - 5 x^{2} - 9 x + 40 - x^{3} - 5 x^{2} - 8 x + 45$

Step 6.2

Simplify by adding terms.

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Step 6.2.1

Combine the opposite terms in $x^{3} - 5 x^{2} - 9 x + 40 - x^{3} - 5 x^{2} - 8 x + 45$ .

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Step 6.2.1.1

Subtract $x^{3}$ from $x^{3}$ .

$L (x) = - 5 x^{2} - 9 x + 40 + 0 - 5 x^{2} - 8 x + 45$

Step 6.2.1.2

Add $- 5 x^{2} - 9 x + 40$ and $0$ .

$L (x) = - 5 x^{2} - 9 x + 40 - 5 x^{2} - 8 x + 45$

Step 6.2.2

Subtract $5 x^{2}$ from $- 5 x^{2}$ .

$L (x) = - 10 x^{2} - 9 x + 40 - 8 x + 45$

Step 6.2.3

Subtract $8 x$ from $- 9 x$ .

$L (x) = - 10 x^{2} - 17 x + 40 + 45$

Step 6.2.4

Add $40$ and $45$ .

$L (x) = - 10 x^{2} - 17 x + 85$

Step 7