Calculus Examples

$f (x) = \tan (x)$ , $a = π$

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 $a = π$ into the linearization function.

$L (x) = f (π) + f' (π) (x - π)$

Step 3

Evaluate $f (π)$ .

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Step 3.1

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

$f (π) = \tan (π)$

Step 3.2

Simplify $\tan (π)$ .

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Step 3.2.1

Remove parentheses.

$\tan (π)$

Step 3.2.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$- \tan (0)$

Step 3.2.3

The exact value of $\tan (0)$ is $0$ .

$- 0$

Step 3.2.4

Multiply $- 1$ by $0$ .

$0$

Step 4

Find the derivative and evaluate it at $π$ .

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Step 4.1

The derivative of $\tan (x)$ with respect to $x$ is $\sec^{2} (x)$ .

$\sec^{2} (x)$

Step 4.2

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

$\sec^{2} (π)$

Step 4.3

Simplify.

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Step 4.3.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

${(- \sec (0))}^{2}$

Step 4.3.2

The exact value of $\sec (0)$ is $1$ .

${(- 1 \cdot 1)}^{2}$

Step 4.3.3

Multiply $- 1$ by $1$ .

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

Step 4.3.4

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

$1$

Step 5

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

$L (x) = 0 + 1 (x - π)$

Step 6

Simplify.

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Step 6.1

Add $0$ and $1 (x - π)$ .

$L (x) = 1 (x - π)$

Step 6.2

Multiply $x - π$ by $1$ .

$L (x) = x - π$

Step 7