Calculus Examples

$f (x) = \cos (x)$

Step 1

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

$- \sin (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (x)]$ .

$f'' (x) = - \frac{d}{d x} \sin (x)$

Step 2.2

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

$f'' (x) = - \cos (x)$

Step 3

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

$- \sin (x) = 0$

Step 4

Divide each term in $- \sin (x) = 0$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- \sin (x) = 0$ by $- 1$ .

$\frac{- \sin (x)}{- 1} = \frac{0}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (x)}{1} = \frac{0}{- 1}$

Step 4.2.2

Divide $\sin (x)$ by $1$ .

$\sin (x) = \frac{0}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 1$ .

$\sin (x) = 0$

Step 5

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$x = 0$

Step 7

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - 0$

Step 8

Subtract $0$ from $π$ .

$x = π$

Step 9

The solution to the equation $x = 0$ .

$x = 0, π$

Step 10

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \cos (0)$

Step 11

Evaluate the second derivative.

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Step 11.1

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

$- 1 \cdot 1$

Step 11.2

Multiply $- 1$ by $1$ .

$- 1$

Step 12

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 13

Find the y-value when $x = 0$ .

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Step 13.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \cos (0)$

Step 13.2

Simplify the result.

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Step 13.2.1

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

$f (0) = 1$

Step 13.2.2

The final answer is $1$ .

$y = 1$

Step 14

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

$- \cos (π)$

Step 15

Evaluate the second derivative.

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Step 15.1

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

$- - \cos (0)$

Step 15.2

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

$- (- 1 \cdot 1)$

Step 15.3

Multiply $- (- 1 \cdot 1)$ .

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Step 15.3.1

Multiply $- 1$ by $1$ .

$- - 1$

Step 15.3.2

Multiply $- 1$ by $- 1$ .

$1$

Step 16

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

$x = π$ is a local minimum

Step 17

Find the y-value when $x = π$ .

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Step 17.1

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

$f (π) = \cos (π)$

Step 17.2

Simplify the result.

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Step 17.2.1

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

$f (π) = - \cos (0)$

Step 17.2.2

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

$f (π) = - 1 \cdot 1$

Step 17.2.3

Multiply $- 1$ by $1$ .

$f (π) = - 1$

Step 17.2.4

The final answer is $- 1$ .

$y = - 1$

Step 18

These are the local extrema for $f (x) = \cos (x)$ .

$(0, 1)$ is a local maxima

$(π, - 1)$ is a local minima

Step 19