Calculus Examples

$y = 1 + 3 \sin (2 (x + \frac{π}{2}))$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $1 + 3 \sin (2 (x + \frac{π}{2}))$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [3 \sin (2 (x + \frac{π}{2}))]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [3 \sin (2 (x + \frac{π}{2}))]$

Step 1.1.2

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

$0 + \frac{d}{d x} [3 \sin (2 (x + \frac{π}{2}))]$

Step 1.2

Evaluate $\frac{d}{d x} [3 \sin (2 (x + \frac{π}{2}))]$ .

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Step 1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \sin (2 (x + \frac{π}{2}))$ with respect to $x$ is $3 \frac{d}{d x} [\sin (2 (x + \frac{π}{2}))]$ .

$0 + 3 \frac{d}{d x} [\sin (2 (x + \frac{π}{2}))]$

Step 1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 (x + \frac{π}{2})$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $2 (x + \frac{π}{2})$ .

$0 + 3 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 (x + \frac{π}{2})])$

Step 1.2.2.2

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

$0 + 3 (\cos (u) \frac{d}{d x} [2 (x + \frac{π}{2})])$

Step 1.2.2.3

Replace all occurrences of $u$ with $2 (x + \frac{π}{2})$ .

$0 + 3 (\cos (2 (x + \frac{π}{2})) \frac{d}{d x} [2 (x + \frac{π}{2})])$

Step 1.2.3

Since $2$ is constant with respect to $x$ , the derivative of $2 (x + \frac{π}{2})$ with respect to $x$ is $2 \frac{d}{d x} [x + \frac{π}{2}]$ .

$0 + 3 (\cos (2 (x + \frac{π}{2})) (2 \frac{d}{d x} [x + \frac{π}{2}]))$

Step 1.2.4

By the Sum Rule, the derivative of $x + \frac{π}{2}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{π}{2}]$ .

$0 + 3 (\cos (2 (x + \frac{π}{2})) (2 (\frac{d}{d x} [x] + \frac{d}{d x} [\frac{π}{2}])))$

Step 1.2.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$0 + 3 (\cos (2 (x + \frac{π}{2})) (2 (1 + \frac{d}{d x} [\frac{π}{2}])))$

Step 1.2.6

Since $\frac{π}{2}$ is constant with respect to $x$ , the derivative of $\frac{π}{2}$ with respect to $x$ is $0$ .

$0 + 3 (\cos (2 (x + \frac{π}{2})) (2 (1 + 0)))$

Step 1.2.7

Add $1$ and $0$ .

$0 + 3 (\cos (2 (x + \frac{π}{2})) (2 \cdot 1))$

Step 1.2.8

Multiply $2$ by $1$ .

$0 + 3 (\cos (2 (x + \frac{π}{2})) \cdot 2)$

Step 1.2.9

Move $2$ to the left of $\cos (2 (x + \frac{π}{2}))$ .

$0 + 3 (2 \cdot \cos (2 (x + \frac{π}{2})))$

Step 1.2.10

Multiply $2$ by $3$ .

$0 + 6 \cos (2 (x + \frac{π}{2}))$

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

$0 + 6 \cos (2 x + 2 \frac{π}{2})$

Step 1.3.2

Combine terms.

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Step 1.3.2.1

Combine $2$ and $\frac{π}{2}$ .

$0 + 6 \cos (2 x + \frac{2 π}{2})$

Step 1.3.2.2

Cancel the common factor of $2$ .

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Step 1.3.2.2.1

Cancel the common factor.

$0 + 6 \cos (2 x + \frac{2 π}{2})$

Step 1.3.2.2.2

Divide $π$ by $1$ .

$0 + 6 \cos (2 x + π)$

Step 1.3.2.3

Add $0$ and $6 \cos (2 x + π)$ .

$6 \cos (2 x + π)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \cos (2 x + π)$ with respect to $x$ is $6 \frac{d}{d x} [\cos (2 x + π)]$ .

$f'' (x) = 6 \frac{d}{d x} (\cos (2 x + π))$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 2 x + π$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $2 x + π$ .

$f'' (x) = 6 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (2 x + π))$

Step 2.2.2

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

$f'' (x) = 6 (- \sin (u) \frac{d}{d x} (2 x + π))$

Step 2.2.3

Replace all occurrences of $u$ with $2 x + π$ .

$f'' (x) = 6 (- \sin (2 x + π) \frac{d}{d x} (2 x + π))$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $6$ .

$f'' (x) = - 6 (\sin (2 x + π) \frac{d}{d x} (2 x + π))$

Step 2.3.2

By the Sum Rule, the derivative of $2 x + π$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [π]$ .

$f'' (x) = - 6 \sin (2 x + π) (\frac{d}{d x} (2 x) + \frac{d}{d x} (π))$

Step 2.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$f'' (x) = - 6 \sin (2 x + π) (2 \frac{d}{d x} (x) + \frac{d}{d x} (π))$

Step 2.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - 6 \sin (2 x + π) (2 \cdot 1 + \frac{d}{d x} (π))$

Step 2.3.5

Multiply $2$ by $1$ .

$f'' (x) = - 6 \sin (2 x + π) (2 + \frac{d}{d x} (π))$

Step 2.3.6

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

$f'' (x) = - 6 \sin (2 x + π) (2 + 0)$

Step 2.3.7

Simplify the expression.

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Step 2.3.7.1

Add $2$ and $0$ .

$f'' (x) = - 6 \sin (2 x + π) \cdot 2$

Step 2.3.7.2

Multiply $2$ by $- 6$ .

$f'' (x) = - 12 \sin (2 x + π)$

Step 3

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

$6 \cos (2 x + π) = 0$

Step 4

Divide each term in $6 \cos (2 x + π) = 0$ by $6$ and simplify.

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Step 4.1

Divide each term in $6 \cos (2 x + π) = 0$ by $6$ .

$\frac{6 \cos (2 x + π)}{6} = \frac{0}{6}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $6$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{6 \cos (2 x + π)}{6} = \frac{0}{6}$

Step 4.2.1.2

Divide $\cos (2 x + π)$ by $1$ .

$\cos (2 x + π) = \frac{0}{6}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $6$ .

$\cos (2 x + π) = 0$

Step 5

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

$2 x + π = \arccos (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$2 x + π = \frac{π}{2}$

Step 7

Move all terms not containing $x$ to the right side of the equation.

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Step 7.1

Subtract $π$ from both sides of the equation.

$2 x = \frac{π}{2} - π$

Step 7.2

To write $- π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 x = \frac{π}{2} - π \cdot \frac{2}{2}$

Step 7.3

Combine $- π$ and $\frac{2}{2}$ .

$2 x = \frac{π}{2} + \frac{- π \cdot 2}{2}$

Step 7.4

Combine the numerators over the common denominator.

$2 x = \frac{π - π \cdot 2}{2}$

Step 7.5

Simplify the numerator.

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Step 7.5.1

Multiply $2$ by $- 1$ .

$2 x = \frac{π - 2 π}{2}$

Step 7.5.2

Subtract $2 π$ from $π$ .

$2 x = \frac{- π}{2}$

Step 7.6

Move the negative in front of the fraction.

$2 x = - \frac{π}{2}$

Step 8

Divide each term in $2 x = - \frac{π}{2}$ by $2$ and simplify.

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Step 8.1

Divide each term in $2 x = - \frac{π}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{- \frac{π}{2}}{2}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $2$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- \frac{π}{2}}{2}$

Step 8.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{2}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{2}$

Step 8.3.2

Multiply $- \frac{π}{2} \cdot \frac{1}{2}$ .

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Step 8.3.2.1

Multiply $\frac{1}{2}$ by $\frac{π}{2}$ .

$x = - \frac{π}{2 \cdot 2}$

Step 8.3.2.2

Multiply $2$ by $2$ .

$x = - \frac{π}{4}$

Step 9

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$2 x + π = 2 π - \frac{π}{2}$

Step 10

Solve for $x$ .

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Step 10.1

Simplify $2 π - \frac{π}{2}$ .

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Step 10.1.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 x + π = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 10.1.2

Combine fractions.

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Step 10.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$2 x + π = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 10.1.2.2

Combine the numerators over the common denominator.

$2 x + π = \frac{2 π \cdot 2 - π}{2}$

Step 10.1.3

Simplify the numerator.

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Step 10.1.3.1

Multiply $2$ by $2$ .

$2 x + π = \frac{4 π - π}{2}$

Step 10.1.3.2

Subtract $π$ from $4 π$ .

$2 x + π = \frac{3 π}{2}$

Step 10.2

Move all terms not containing $x$ to the right side of the equation.

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Step 10.2.1

Subtract $π$ from both sides of the equation.

$2 x = \frac{3 π}{2} - π$

Step 10.2.2

To write $- π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 x = \frac{3 π}{2} - π \cdot \frac{2}{2}$

Step 10.2.3

Combine $- π$ and $\frac{2}{2}$ .

$2 x = \frac{3 π}{2} + \frac{- π \cdot 2}{2}$

Step 10.2.4

Combine the numerators over the common denominator.

$2 x = \frac{3 π - π \cdot 2}{2}$

Step 10.2.5

Simplify the numerator.

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Step 10.2.5.1

Multiply $2$ by $- 1$ .

$2 x = \frac{3 π - 2 π}{2}$

Step 10.2.5.2

Subtract $2 π$ from $3 π$ .

$2 x = \frac{π}{2}$

Step 10.3

Divide each term in $2 x = \frac{π}{2}$ by $2$ and simplify.

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Step 10.3.1

Divide each term in $2 x = \frac{π}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{π}{2}}{2}$

Step 10.3.2

Simplify the left side.

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Step 10.3.2.1

Cancel the common factor of $2$ .

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Step 10.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{π}{2}}{2}$

Step 10.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{2}$

Step 10.3.3

Simplify the right side.

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Step 10.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{2}$

Step 10.3.3.2

Multiply $\frac{π}{2} \cdot \frac{1}{2}$ .

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Step 10.3.3.2.1

Multiply $\frac{π}{2}$ by $\frac{1}{2}$ .

$x = \frac{π}{2 \cdot 2}$

Step 10.3.3.2.2

Multiply $2$ by $2$ .

$x = \frac{π}{4}$

Step 11

The solution to the equation $2 x + π = \frac{π}{2}$ .

$x = - \frac{π}{4}, \frac{π}{4}$

Step 12

Evaluate the second derivative at $x = - \frac{π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 \sin (2 (- \frac{π}{4}) + π)$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

Cancel the common factor of $2$ .

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Step 13.1.1.1

Move the leading negative in $- \frac{π}{4}$ into the numerator.

$- 12 \sin (2 \frac{- π}{4} + π)$

Step 13.1.1.2

Factor $2$ out of $4$ .

$- 12 \sin (2 \frac{- π}{2 (2)} + π)$

Step 13.1.1.3

Cancel the common factor.

$- 12 \sin (2 \frac{- π}{2 \cdot 2} + π)$

Step 13.1.1.4

Rewrite the expression.

$- 12 \sin (\frac{- π}{2} + π)$

Step 13.1.2

Move the negative in front of the fraction.

$- 12 \sin (- \frac{π}{2} + π)$

Step 13.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 12 \sin (- \frac{π}{2} + π \cdot \frac{2}{2})$

Step 13.3

Combine fractions.

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Step 13.3.1

Combine $π$ and $\frac{2}{2}$ .

$- 12 \sin (- \frac{π}{2} + \frac{π \cdot 2}{2})$

Step 13.3.2

Combine the numerators over the common denominator.

$- 12 \sin (\frac{- π + π \cdot 2}{2})$

Step 13.4

Simplify the numerator.

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Step 13.4.1

Move $2$ to the left of $π$ .

$- 12 \sin (\frac{- π + 2 \cdot π}{2})$

Step 13.4.2

Add $- π$ and $2 π$ .

$- 12 \sin (\frac{π}{2})$

Step 13.5

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

$- 12 \cdot 1$

Step 13.6

Multiply $- 12$ by $1$ .

$- 12$

Step 14

$x = - \frac{π}{4}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - \frac{π}{4}$ is a local maximum

Step 15

Find the y-value when $x = - \frac{π}{4}$ .

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Step 15.1

Replace the variable $x$ with $- \frac{π}{4}$ in the expression.

$f (- \frac{π}{4}) = 1 + 3 \sin (2 ((- \frac{π}{4}) + \frac{π}{2}))$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

To write $\frac{π}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (- \frac{π}{4} + \frac{π}{2} \cdot \frac{2}{2}))$

Step 15.2.1.2

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 15.2.1.2.1

Multiply $\frac{π}{2}$ by $\frac{2}{2}$ .

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (- \frac{π}{4} + \frac{π \cdot 2}{2 \cdot 2}))$

Step 15.2.1.2.2

Multiply $2$ by $2$ .

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (- \frac{π}{4} + \frac{π \cdot 2}{4}))$

Step 15.2.1.3

Combine the numerators over the common denominator.

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (\frac{- π + π \cdot 2}{4}))$

Step 15.2.1.4

Simplify the numerator.

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Step 15.2.1.4.1

Move $2$ to the left of $π$ .

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (\frac{- π + 2 \cdot π}{4}))$

Step 15.2.1.4.2

Add $- π$ and $2 π$ .

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (\frac{π}{4}))$

Step 15.2.1.5

Cancel the common factor of $2$ .

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Step 15.2.1.5.1

Factor $2$ out of $4$ .

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (\frac{π}{2 (2)}))$

Step 15.2.1.5.2

Cancel the common factor.

$f (- \frac{π}{4}) = 1 + 3 \sin (2 (\frac{π}{2 \cdot 2}))$

Step 15.2.1.5.3

Rewrite the expression.

$f (- \frac{π}{4}) = 1 + 3 \sin (\frac{π}{2})$

Step 15.2.1.6

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

$f (- \frac{π}{4}) = 1 + 3 \cdot 1$

Step 15.2.1.7

Multiply $3$ by $1$ .

$f (- \frac{π}{4}) = 1 + 3$

Step 15.2.2

Add $1$ and $3$ .

$f (- \frac{π}{4}) = 4$

Step 15.2.3

The final answer is $4$ .

$y = 4$

Step 16

Evaluate the second derivative at $x = \frac{π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 \sin (2 (\frac{π}{4}) + π)$

Step 17

Evaluate the second derivative.

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Step 17.1

Cancel the common factor of $2$ .

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Step 17.1.1

Factor $2$ out of $4$ .

$- 12 \sin (2 \frac{π}{2 (2)} + π)$

Step 17.1.2

Cancel the common factor.

$- 12 \sin (2 \frac{π}{2 \cdot 2} + π)$

Step 17.1.3

Rewrite the expression.

$- 12 \sin (\frac{π}{2} + π)$

Step 17.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 12 \sin (\frac{π}{2} + π \cdot \frac{2}{2})$

Step 17.3

Combine fractions.

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Step 17.3.1

Combine $π$ and $\frac{2}{2}$ .

$- 12 \sin (\frac{π}{2} + \frac{π \cdot 2}{2})$

Step 17.3.2

Combine the numerators over the common denominator.

$- 12 \sin (\frac{π + π \cdot 2}{2})$

Step 17.4

Simplify the numerator.

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Step 17.4.1

Move $2$ to the left of $π$ .

$- 12 \sin (\frac{π + 2 \cdot π}{2})$

Step 17.4.2

Add $π$ and $2 π$ .

$- 12 \sin (\frac{3 π}{2})$

Step 17.5

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

$- 12 (- \sin (\frac{π}{2}))$

Step 17.6

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

$- 12 (- 1 \cdot 1)$

Step 17.7

Multiply $- 12 (- 1 \cdot 1)$ .

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Step 17.7.1

Multiply $- 1$ by $1$ .

$- 12 \cdot - 1$

Step 17.7.2

Multiply $- 12$ by $- 1$ .

$12$

Step 18

$x = \frac{π}{4}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{π}{4}$ is a local minimum

Step 19

Find the y-value when $x = \frac{π}{4}$ .

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Step 19.1

Replace the variable $x$ with $\frac{π}{4}$ in the expression.

$f (\frac{π}{4}) = 1 + 3 \sin (2 ((\frac{π}{4}) + \frac{π}{2}))$

Step 19.2

Simplify the result.

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Step 19.2.1

Simplify each term.

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Step 19.2.1.1

To write $\frac{π}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{π}{4} + \frac{π}{2} \cdot \frac{2}{2}))$

Step 19.2.1.2

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 19.2.1.2.1

Multiply $\frac{π}{2}$ by $\frac{2}{2}$ .

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{π}{4} + \frac{π \cdot 2}{2 \cdot 2}))$

Step 19.2.1.2.2

Multiply $2$ by $2$ .

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{π}{4} + \frac{π \cdot 2}{4}))$

Step 19.2.1.3

Combine the numerators over the common denominator.

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{π + π \cdot 2}{4}))$

Step 19.2.1.4

Simplify the numerator.

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Step 19.2.1.4.1

Move $2$ to the left of $π$ .

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{π + 2 \cdot π}{4}))$

Step 19.2.1.4.2

Add $π$ and $2 π$ .

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{3 π}{4}))$

Step 19.2.1.5

Cancel the common factor of $2$ .

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Step 19.2.1.5.1

Factor $2$ out of $4$ .

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{3 π}{2 (2)}))$

Step 19.2.1.5.2

Cancel the common factor.

$f (\frac{π}{4}) = 1 + 3 \sin (2 (\frac{3 π}{2 \cdot 2}))$

Step 19.2.1.5.3

Rewrite the expression.

$f (\frac{π}{4}) = 1 + 3 \sin (\frac{3 π}{2})$

Step 19.2.1.6

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

$f (\frac{π}{4}) = 1 + 3 (- \sin (\frac{π}{2}))$

Step 19.2.1.7

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

$f (\frac{π}{4}) = 1 + 3 (- 1 \cdot 1)$

Step 19.2.1.8

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

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Step 19.2.1.8.1

Multiply $- 1$ by $1$ .

$f (\frac{π}{4}) = 1 + 3 \cdot - 1$

Step 19.2.1.8.2

Multiply $3$ by $- 1$ .

$f (\frac{π}{4}) = 1 - 3$

Step 19.2.2

Subtract $3$ from $1$ .

$f (\frac{π}{4}) = - 2$

Step 19.2.3

The final answer is $- 2$ .

$y = - 2$

Step 20

These are the local extrema for $f (x) = 1 + 3 \sin (2 (x + \frac{π}{2}))$ .

$(- \frac{π}{4}, 4)$ is a local maxima

$(\frac{π}{4}, - 2)$ is a local minima

Step 21