Calculus Examples

$y = 3 \cos (x) + 1$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

By the Sum Rule, the derivative of $3 \cos (x) + 1$ with respect to $x$ is $\frac{d}{d x} [3 \cos (x)] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [3 \cos (x)] + \frac{d}{d x} [1]$

Step 1.2

Evaluate $\frac{d}{d x} [3 \cos (x)]$ .

Tap for more steps...

Step 1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \cos (x)$ with respect to $x$ is $3 \frac{d}{d x} [\cos (x)]$ .

$3 \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [1]$

Step 1.2.2

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

$3 (- \sin (x)) + \frac{d}{d x} [1]$

Step 1.2.3

Multiply $- 1$ by $3$ .

$- 3 \sin (x) + \frac{d}{d x} [1]$

Step 1.3

Differentiate using the Constant Rule.

Tap for more steps...

Step 1.3.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$- 3 \sin (x) + 0$

Step 1.3.2

Add $- 3 \sin (x)$ and $0$ .

$- 3 \sin (x)$

Step 2

Find the second derivative of the function.

Tap for more steps...

Step 2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 \sin (x)$ with respect to $x$ is $- 3 \frac{d}{d x} [\sin (x)]$ .

$f'' (x) = - 3 \frac{d}{d x} \sin (x)$

Step 2.2

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

$f'' (x) = - 3 \cos (x)$

Step 3

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

$- 3 \sin (x) = 0$

Step 4

Divide each term in $- 3 \sin (x) = 0$ by $- 3$ and simplify.

Tap for more steps...

Step 4.1

Divide each term in $- 3 \sin (x) = 0$ by $- 3$ .

$\frac{- 3 \sin (x)}{- 3} = \frac{0}{- 3}$

Step 4.2

Simplify the left side.

Tap for more steps...

Step 4.2.1

Cancel the common factor of $- 3$ .

Tap for more steps...

Step 4.2.1.1

Cancel the common factor.

$\frac{- 3 \sin (x)}{- 3} = \frac{0}{- 3}$

Step 4.2.1.2

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

$\sin (x) = \frac{0}{- 3}$

Step 4.3

Simplify the right side.

Tap for more steps...

Step 4.3.1

Divide $0$ by $- 3$ .

$\sin (x) = 0$

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.

Tap for more steps...

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.

$- 3 \cos (0)$

Step 11

Evaluate the second derivative.

Tap for more steps...

Step 11.1

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

$- 3 \cdot 1$

Step 11.2

Multiply $- 3$ by $1$ .

$- 3$

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 $x = 0$ .

Tap for more steps...

Step 13.1

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

$f (0) = 3 \cos (0) + 1$

Step 13.2

Simplify the result.

Tap for more steps...

Step 13.2.1

Simplify each term.

Tap for more steps...

Step 13.2.1.1

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

$f (0) = 3 \cdot 1 + 1$

Step 13.2.1.2

Multiply $3$ by $1$ .

$f (0) = 3 + 1$

Step 13.2.2

Add $3$ and $1$ .

$f (0) = 4$

Step 13.2.3

The final answer is $4$ .

$y = 4$

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.

$- 3 \cos (π)$

Step 15

Evaluate the second derivative.

Tap for more steps...

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.

$- 3 (- \cos (0))$

Step 15.2

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

$- 3 (- 1 \cdot 1)$

Step 15.3

Multiply $- 3 (- 1 \cdot 1)$ .

Tap for more steps...

Step 15.3.1

Multiply $- 1$ by $1$ .

$- 3 \cdot - 1$

Step 15.3.2

Multiply $- 3$ by $- 1$ .

$3$

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 $x = π$ .

Tap for more steps...

Step 17.1

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

$f (π) = 3 \cos (π) + 1$

Step 17.2

Simplify the result.

Tap for more steps...

Step 17.2.1

Simplify each term.

Tap for more steps...

Step 17.2.1.1

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.

$f (π) = 3 (- \cos (0)) + 1$

Step 17.2.1.2

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

$f (π) = 3 (- 1 \cdot 1) + 1$

Step 17.2.1.3

Multiply $3 (- 1 \cdot 1)$ .

Tap for more steps...

Step 17.2.1.3.1

Multiply $- 1$ by $1$ .

$f (π) = 3 \cdot - 1 + 1$

Step 17.2.1.3.2

Multiply $3$ by $- 1$ .

$f (π) = - 3 + 1$

Step 17.2.2

Add $- 3$ and $1$ .

$f (π) = - 2$

Step 17.2.3

The final answer is $- 2$ .

$y = - 2$

Step 18

These are the local extrema for $f (x) = 3 \cos (x) + 1$ .

$(0, 4)$ is a local maxima

$(π, - 2)$ is a local minima

Step 19