Calculus Examples

$f (x) = π \cos (π x) + 1$

Step 1

Use the form $a \cos (b x - c) + d$ to find the variables used to find the amplitude, period, phase shift, and vertical shift.

$a = π$

$b = π$

$c = 0$

$d = 1$

Step 2

Find the amplitude $| a |$ .

Amplitude: $π$

Step 3

Find the period using the formula $\frac{2 π}{| b |}$ .

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Step 3.1

Find the period of $π \cos (π x)$ .

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Step 3.1.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.1.2

Replace $b$ with $π$ in the formula for period.

$\frac{2 π}{| π |}$

Step 3.1.3

$π$ is approximately $3.14159265$ which is positive so remove the absolute value

$\frac{2 π}{π}$

Step 3.1.4

Cancel the common factor of $π$ .

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Step 3.1.4.1

Cancel the common factor.

$\frac{2 π}{π}$

Step 3.1.4.2

Divide $2$ by $1$ .

$2$

Step 3.2

Find the period of $1$ .

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Step 3.2.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.2.2

Replace $b$ with $π$ in the formula for period.

$\frac{2 π}{| π |}$

Step 3.2.3

$π$ is approximately $3.14159265$ which is positive so remove the absolute value

$\frac{2 π}{π}$

Step 3.2.4

Cancel the common factor of $π$ .

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Step 3.2.4.1

Cancel the common factor.

$\frac{2 π}{π}$

Step 3.2.4.2

Divide $2$ by $1$ .

$2$

Step 3.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

$2$

Step 4

Find the phase shift using the formula $\frac{c}{b}$ .

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Step 4.1

The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Step 4.2

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{0}{π}$

Step 4.3

Divide $0$ by $π$ .

Phase Shift: $0$

Step 5

List the properties of the trigonometric function.

Amplitude: $π$

Period: $2$

Phase Shift: None

Vertical Shift: $1$

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = 0$ .

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Step 6.1.1

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

$f (0) = π \cos (π (0)) + 1$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Simplify each term.

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Step 6.1.2.1.1

Multiply $π$ by $0$ .

$f (0) = π \cos (0) + 1$

Step 6.1.2.1.2

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

$f (0) = π \cdot 1 + 1$

Step 6.1.2.1.3

Multiply $π$ by $1$ .

$f (0) = π + 1$

Step 6.1.2.2

The final answer is $π + 1$ .

$π + 1$

Step 6.2

Find the point at $x = \frac{1}{2}$ .

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Step 6.2.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = π \cos (π (\frac{1}{2})) + 1$

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Combine $π$ and $\frac{1}{2}$ .

$f (\frac{1}{2}) = π \cos (\frac{π}{2}) + 1$

Step 6.2.2.1.2

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

$f (\frac{1}{2}) = π \cdot 0 + 1$

Step 6.2.2.1.3

Multiply $π$ by $0$ .

$f (\frac{1}{2}) = 0 + 1$

Step 6.2.2.2

Add $0$ and $1$ .

$f (\frac{1}{2}) = 1$

Step 6.2.2.3

The final answer is $1$ .

$1$

Step 6.3

Find the point at $x = 1$ .

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Step 6.3.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = π \cos (π (1)) + 1$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Simplify each term.

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Step 6.3.2.1.1

Multiply $π$ by $1$ .

$f (1) = π \cos (π) + 1$

Step 6.3.2.1.2

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.

$f (1) = π (- \cos (0)) + 1$

Step 6.3.2.1.3

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

$f (1) = π (- 1 \cdot 1) + 1$

Step 6.3.2.1.4

Multiply $- 1$ by $1$ .

$f (1) = π \cdot - 1 + 1$

Step 6.3.2.1.5

Move $- 1$ to the left of $π$ .

$f (1) = - 1 \cdot π + 1$

Step 6.3.2.1.6

Rewrite $- 1 π$ as $- π$ .

$f (1) = - π + 1$

Step 6.3.2.2

The final answer is $- π + 1$ .

$- π + 1$

Step 6.4

Find the point at $x = \frac{3}{2}$ .

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Step 6.4.1

Replace the variable $x$ with $\frac{3}{2}$ in the expression.

$f (\frac{3}{2}) = π \cos (π (\frac{3}{2})) + 1$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Simplify each term.

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Step 6.4.2.1.1

Combine $π$ and $\frac{3}{2}$ .

$f (\frac{3}{2}) = π \cos (\frac{π \cdot 3}{2}) + 1$

Step 6.4.2.1.2

Move $3$ to the left of $π$ .

$f (\frac{3}{2}) = π \cos (\frac{3 \cdot π}{2}) + 1$

Step 6.4.2.1.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (\frac{3}{2}) = π \cos (\frac{π}{2}) + 1$

Step 6.4.2.1.4

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

$f (\frac{3}{2}) = π \cdot 0 + 1$

Step 6.4.2.1.5

Multiply $π$ by $0$ .

$f (\frac{3}{2}) = 0 + 1$

Step 6.4.2.2

Add $0$ and $1$ .

$f (\frac{3}{2}) = 1$

Step 6.4.2.3

The final answer is $1$ .

$1$

Step 6.5

Find the point at $x = 2$ .

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Step 6.5.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = π \cos (π (2)) + 1$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Simplify each term.

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Step 6.5.2.1.1

Move $2$ to the left of $π$ .

$f (2) = π \cos (2 \cdot π) + 1$

Step 6.5.2.1.2

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

$f (2) = π \cos (0) + 1$

Step 6.5.2.1.3

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

$f (2) = π \cdot 1 + 1$

Step 6.5.2.1.4

Multiply $π$ by $1$ .

$f (2) = π + 1$

Step 6.5.2.2

The final answer is $π + 1$ .

$π + 1$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & π + 1 \\ \frac{1}{2} & 1 \\ 1 & - π + 1 \\ \frac{3}{2} & 1 \\ 2 & π + 1 \end{array}$

Step 7

The trig function can be graphed using the amplitude, period, phase shift, vertical shift, and the points.

Amplitude: $π$

Period: $2$

Phase Shift: None

Vertical Shift: $1$

$\begin{array}{c} x & f (x) \\ 0 & π + 1 \\ \frac{1}{2} & 1 \\ 1 & - π + 1 \\ \frac{3}{2} & 1 \\ 2 & π + 1 \end{array}$

Step 8