Calculus Examples

$f (x) = \cos (x)$ , $a = \frac{5 π}{2}$

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 $a = \frac{5 π}{2}$ into the linearization function.

$L (x) = f (\frac{5 π}{2}) + f' (\frac{5 π}{2}) (x - \frac{5 π}{2})$

Step 3

Evaluate $f (\frac{5 π}{2})$ .

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Step 3.1

Replace the variable $x$ with $\frac{5 π}{2}$ in the expression.

$f (\frac{5 π}{2}) = \cos (\frac{5 π}{2})$

Step 3.2

Simplify $\cos (\frac{5 π}{2})$ .

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Step 3.2.1

Remove parentheses.

$\cos (\frac{5 π}{2})$

Step 3.2.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\cos (\frac{π}{2})$

Step 3.2.3

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$0$

Step 4

Find the derivative and evaluate it at $\frac{5 π}{2}$ .

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Step 4.1

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$- \sin (x)$

Step 4.2

Replace the variable $x$ with $\frac{5 π}{2}$ in the expression.

$- \sin (\frac{5 π}{2})$

Step 4.3

Simplify.

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Step 4.3.1

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \sin (\frac{π}{2})$

Step 4.3.2

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$- 1 \cdot 1$

Step 4.3.3

Multiply $- 1$ by $1$ .

$- 1$

Step 5

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

$L (x) = 0 - (x - \frac{5 π}{2})$

Step 6

Simplify.

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Step 6.1

Subtract $x - \frac{5 π}{2}$ from $0$ .

$L (x) = - (x - \frac{5 π}{2})$

Step 6.2

Apply the distributive property.

$L (x) = - x + \frac{5 π}{2}$

Step 6.3

Multiply $- - \frac{5 π}{2}$ .

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Step 6.3.1

Multiply $- 1$ by $- 1$ .

$L (x) = - x + 1 (\frac{5 π}{2})$

Step 6.3.2

Multiply $\frac{5 π}{2}$ by $1$ .

$L (x) = - x + \frac{5 π}{2}$

Step 7