Calculus Examples

$y = 2 \sin (x) - \cos^{2} (x)$

Step 1

Write $y = 2 \sin (x) - \cos^{2} (x)$ as a function.

$f (x) = 2 \sin (x) - \cos^{2} (x)$

Step 2

Find the first derivative of the function.

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Step 2.1

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

$\frac{d}{d x} [2 \sin (x)] + \frac{d}{d x} [- \cos^{2} (x)]$

Step 2.2

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

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Step 2.2.1

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

$2 \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [- \cos^{2} (x)]$

Step 2.2.2

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

$2 \cos (x) + \frac{d}{d x} [- \cos^{2} (x)]$

Step 2.3

Evaluate $\frac{d}{d x} [- \cos^{2} (x)]$ .

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Step 2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos^{2} (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos^{2} (x)]$ .

$2 \cos (x) - \frac{d}{d x} [\cos^{2} (x)]$

Step 2.3.2

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

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Step 2.3.2.1

To apply the Chain Rule, set $u$ as $\cos (x)$ .

$2 \cos (x) - (\frac{d}{d u} [u^{2}] \frac{d}{d x} [\cos (x)])$

Step 2.3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$2 \cos (x) - (2 u \frac{d}{d x} [\cos (x)])$

Step 2.3.2.3

Replace all occurrences of $u$ with $\cos (x)$ .

$2 \cos (x) - (2 \cos (x) \frac{d}{d x} [\cos (x)])$

Step 2.3.3

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

$2 \cos (x) - (2 \cos (x) (- \sin (x)))$

Step 2.3.4

Multiply $- 1$ by $2$ .

$2 \cos (x) - (- 2 \cos (x) \sin (x))$

Step 2.3.5

Multiply $- 2$ by $- 1$ .

$2 \cos (x) + 2 \cos (x) \sin (x)$

Step 2.4

Simplify.

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Step 2.4.1

Reorder terms.

$2 \cos (x) \sin (x) + 2 \cos (x)$

Step 2.4.2

Simplify each term.

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Step 2.4.2.1

Reorder $2 \cos (x)$ and $\sin (x)$ .

$\sin (x) (2 \cos (x)) + 2 \cos (x)$

Step 2.4.2.2

Reorder $\sin (x)$ and $2$ .

$2 \cdot \sin (x) \cos (x) + 2 \cos (x)$

Step 2.4.2.3

Apply the sine double-angle identity.

$\sin (2 x) + 2 \cos (x)$

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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 $g (x) = 2 x$ .

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Step 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.2

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

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

Step 3.2.3

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

$f'' (x) = \cos (2 x) (2 \cdot 1) + \frac{d}{d x} (2 \cos (x))$

Step 3.2.4

Multiply $2$ by $1$ .

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

Step 3.2.5

Move $2$ to the left of $\cos (2 x)$ .

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

Step 3.3

Evaluate $\frac{d}{d x} [2 \cos (x)]$ .

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Step 3.3.1

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

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

Step 3.3.2

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

$f'' (x) = 2 \cos (2 x) + 2 (- \sin (x))$

Step 3.3.3

Multiply $- 1$ by $2$ .

$f'' (x) = 2 \cos (2 x) - 2 \sin (x)$

Step 4

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

$\sin (2 x) + 2 \cos (x) = 0$

Step 5

Apply the sine double-angle identity.

$2 \sin (x) \cos (x) + 2 \cos (x) = 0$

Step 6

Factor $2 \cos (x)$ out of $2 \sin (x) \cos (x) + 2 \cos (x)$ .

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Step 6.1

Factor $2 \cos (x)$ out of $2 \sin (x) \cos (x)$ .

$2 \cos (x) \sin (x) + 2 \cos (x) = 0$

Step 6.2

Factor $2 \cos (x)$ out of $2 \cos (x)$ .

$2 \cos (x) \sin (x) + 2 \cos (x) \cdot 1 = 0$

Step 6.3

Factor $2 \cos (x)$ out of $2 \cos (x) \sin (x) + 2 \cos (x) \cdot 1$ .

$2 \cos (x) (\sin (x) + 1) = 0$

Step 7

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (x) = 0$

$\sin (x) + 1 = 0$

Step 8

Set $\cos (x)$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Step 8.2

Solve $\cos (x) = 0$ for $x$ .

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Step 8.2.1

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

$x = \arccos (0)$

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

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

$x = \frac{π}{2}$

Step 8.2.3

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.

$x = 2 π - \frac{π}{2}$

Step 8.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 8.2.4.1

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

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

Step 8.2.4.2

Combine fractions.

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Step 8.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 8.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 8.2.4.3

Simplify the numerator.

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Step 8.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 8.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 8.2.5

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

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 9

Set $\sin (x) + 1$ equal to $0$ and solve for $x$ .

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Step 9.1

Set $\sin (x) + 1$ equal to $0$ .

$\sin (x) + 1 = 0$

Step 9.2

Solve $\sin (x) + 1 = 0$ for $x$ .

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Step 9.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 9.2.2

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

$x = \arcsin (- 1)$

Step 9.2.3

Simplify the right side.

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Step 9.2.3.1

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

$x = - \frac{π}{2}$

Step 9.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 9.2.5

Simplify the expression to find the second solution.

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Step 9.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 9.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 9.2.6

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

$x = - \frac{π}{2}, \frac{3 π}{2}$

Step 10

The final solution is all the values that make $2 \cos (x) (\sin (x) + 1) = 0$ true.

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

Step 11

Evaluate the second derivative at $x = \frac{π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \cos (2 (\frac{π}{2})) - 2 \sin (\frac{π}{2})$

Step 12

Evaluate the second derivative.

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Step 12.1

Simplify each term.

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Step 12.1.1

Cancel the common factor of $2$ .

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Step 12.1.1.1

Cancel the common factor.

$2 \cos (2 \frac{π}{2}) - 2 \sin (\frac{π}{2})$

Step 12.1.1.2

Rewrite the expression.

$2 \cos (π) - 2 \sin (\frac{π}{2})$

Step 12.1.2

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.

$2 (- \cos (0)) - 2 \sin (\frac{π}{2})$

Step 12.1.3

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

$2 (- 1 \cdot 1) - 2 \sin (\frac{π}{2})$

Step 12.1.4

Multiply $2 (- 1 \cdot 1)$ .

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Step 12.1.4.1

Multiply $- 1$ by $1$ .

$2 \cdot - 1 - 2 \sin (\frac{π}{2})$

Step 12.1.4.2

Multiply $2$ by $- 1$ .

$- 2 - 2 \sin (\frac{π}{2})$

Step 12.1.5

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

$- 2 - 2 \cdot 1$

Step 12.1.6

Multiply $- 2$ by $1$ .

$- 2 - 2$

Step 12.2

Subtract $2$ from $- 2$ .

$- 4$

Step 13

$x = \frac{π}{2}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{2}$ is a local maximum

Step 14

Find the y-value when $x = \frac{π}{2}$ .

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Step 14.1

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

$f (\frac{π}{2}) = 2 \sin (\frac{π}{2}) - \cos^{2} (\frac{π}{2})$

Step 14.2

Simplify the result.

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Step 14.2.1

Simplify each term.

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Step 14.2.1.1

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

$f (\frac{π}{2}) = 2 \cdot 1 - \cos^{2} (\frac{π}{2})$

Step 14.2.1.2

Multiply $2$ by $1$ .

$f (\frac{π}{2}) = 2 - \cos^{2} (\frac{π}{2})$

Step 14.2.1.3

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

$f (\frac{π}{2}) = 2 - 0^{2}$

Step 14.2.1.4

Raising $0$ to any positive power yields $0$ .

$f (\frac{π}{2}) = 2 - 0$

Step 14.2.1.5

Multiply $- 1$ by $0$ .

$f (\frac{π}{2}) = 2 + 0$

Step 14.2.2

Add $2$ and $0$ .

$f (\frac{π}{2}) = 2$

Step 14.2.3

The final answer is $2$ .

$y = 2$

Step 15

Evaluate the second derivative at $x = \frac{3 π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \cos (2 (\frac{3 π}{2})) - 2 \sin (\frac{3 π}{2})$

Step 16

Evaluate the second derivative.

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Step 16.1

Simplify each term.

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Step 16.1.1

Cancel the common factor of $2$ .

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Step 16.1.1.1

Cancel the common factor.

$2 \cos (2 \frac{3 π}{2}) - 2 \sin (\frac{3 π}{2})$

Step 16.1.1.2

Rewrite the expression.

$2 \cos (3 π) - 2 \sin (\frac{3 π}{2})$

Step 16.1.2

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

$2 \cos (π) - 2 \sin (\frac{3 π}{2})$

Step 16.1.3

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.

$2 (- \cos (0)) - 2 \sin (\frac{3 π}{2})$

Step 16.1.4

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

$2 (- 1 \cdot 1) - 2 \sin (\frac{3 π}{2})$

Step 16.1.5

Multiply $2 (- 1 \cdot 1)$ .

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Step 16.1.5.1

Multiply $- 1$ by $1$ .

$2 \cdot - 1 - 2 \sin (\frac{3 π}{2})$

Step 16.1.5.2

Multiply $2$ by $- 1$ .

$- 2 - 2 \sin (\frac{3 π}{2})$

Step 16.1.6

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.

$- 2 - 2 (- \sin (\frac{π}{2}))$

Step 16.1.7

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

$- 2 - 2 (- 1 \cdot 1)$

Step 16.1.8

Multiply $- 2 (- 1 \cdot 1)$ .

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Step 16.1.8.1

Multiply $- 1$ by $1$ .

$- 2 - 2 \cdot - 1$

Step 16.1.8.2

Multiply $- 2$ by $- 1$ .

$- 2 + 2$

Step 16.2

Add $- 2$ and $2$ .

$0$

Step 17

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 17.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{2}) \cup (\frac{3 π}{2}, \infty)$

Step 17.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - \frac{π}{2})$ in the first derivative $\sin (2 x) + 2 \cos (x)$ to check if the result is negative or positive.

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Step 17.2.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = \sin (2 (- 4)) + 2 \cos (- 4)$

Step 17.2.2

Simplify the result.

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Step 17.2.2.1

Simplify each term.

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Step 17.2.2.1.1

Multiply $2$ by $- 4$ .

$f' (- 4) = \sin (- 8) + 2 \cos (- 4)$

Step 17.2.2.1.2

Evaluate $\sin (- 8)$ .

$f' (- 4) = - 0.98935824 + 2 \cos (- 4)$

Step 17.2.2.1.3

Evaluate $\cos (- 4)$ .

$f' (- 4) = - 0.98935824 + 2 \cdot - 0.65364362$

Step 17.2.2.1.4

Multiply $2$ by $- 0.65364362$ .

$f' (- 4) = - 0.98935824 - 1.30728724$

Step 17.2.2.2

Subtract $1.30728724$ from $- 0.98935824$ .

$f' (- 4) = - 2.29664548$

Step 17.2.2.3

The final answer is $- 2.29664548$ .

$- 2.29664548$

Step 17.3

Substitute any number, such as $0$ , from the interval $(- \frac{π}{2}, \frac{π}{2})$ in the first derivative $\sin (2 x) + 2 \cos (x)$ to check if the result is negative or positive.

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Step 17.3.1

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

$f' (0) = \sin (2 (0)) + 2 \cos (0)$

Step 17.3.2

Simplify the result.

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Step 17.3.2.1

Simplify each term.

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Step 17.3.2.1.1

Multiply $2$ by $0$ .

$f' (0) = \sin (0) + 2 \cos (0)$

Step 17.3.2.1.2

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

$f' (0) = 0 + 2 \cos (0)$

Step 17.3.2.1.3

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

$f' (0) = 0 + 2 \cdot 1$

Step 17.3.2.1.4

Multiply $2$ by $1$ .

$f' (0) = 0 + 2$

Step 17.3.2.2

Add $0$ and $2$ .

$f' (0) = 2$

Step 17.3.2.3

The final answer is $2$ .

$2$

Step 17.4

Substitute any number, such as $3$ , from the interval $(\frac{π}{2}, \frac{3 π}{2})$ in the first derivative $\sin (2 x) + 2 \cos (x)$ to check if the result is negative or positive.

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Step 17.4.1

Replace the variable $x$ with $3$ in the expression.

$f' (3) = \sin (2 (3)) + 2 \cos (3)$

Step 17.4.2

Simplify the result.

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Step 17.4.2.1

Simplify each term.

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Step 17.4.2.1.1

Multiply $2$ by $3$ .

$f' (3) = \sin (6) + 2 \cos (3)$

Step 17.4.2.1.2

Evaluate $\sin (6)$ .

$f' (3) = - 0.27941549 + 2 \cos (3)$

Step 17.4.2.1.3

Evaluate $\cos (3)$ .

$f' (3) = - 0.27941549 + 2 \cdot - 0.98999249$

Step 17.4.2.1.4

Multiply $2$ by $- 0.98999249$ .

$f' (3) = - 0.27941549 - 1.97998499$

Step 17.4.2.2

Subtract $1.97998499$ from $- 0.27941549$ .

$f' (3) = - 2.25940049$

Step 17.4.2.3

The final answer is $- 2.25940049$ .

$- 2.25940049$

Step 17.5

Substitute any number, such as $7$ , from the interval $(\frac{3 π}{2}, \infty)$ in the first derivative $\sin (2 x) + 2 \cos (x)$ to check if the result is negative or positive.

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Step 17.5.1

Replace the variable $x$ with $7$ in the expression.

$f' (7) = \sin (2 (7)) + 2 \cos (7)$

Step 17.5.2

Simplify the result.

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Step 17.5.2.1

Simplify each term.

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Step 17.5.2.1.1

Multiply $2$ by $7$ .

$f' (7) = \sin (14) + 2 \cos (7)$

Step 17.5.2.1.2

Evaluate $\sin (14)$ .

$f' (7) = 0.99060735 + 2 \cos (7)$

Step 17.5.2.1.3

Evaluate $\cos (7)$ .

$f' (7) = 0.99060735 + 2 \cdot 0.75390225$

Step 17.5.2.1.4

Multiply $2$ by $0.75390225$ .

$f' (7) = 0.99060735 + 1.5078045$

Step 17.5.2.2

Add $0.99060735$ and $1.5078045$ .

$f' (7) = 2.49841186$

Step 17.5.2.3

The final answer is $2.49841186$ .

$2.49841186$

Step 17.6

Since the first derivative changed signs from negative to positive around $x = - \frac{π}{2}$ , then $x = - \frac{π}{2}$ is a local minimum.

$x = - \frac{π}{2}$ is a local minimum

Step 17.7

Since the first derivative changed signs from positive to negative around $x = \frac{π}{2}$ , then $x = \frac{π}{2}$ is a local maximum.

$x = \frac{π}{2}$ is a local maximum

Step 17.8

Since the first derivative changed signs from negative to positive around $x = \frac{3 π}{2}$ , then $x = \frac{3 π}{2}$ is a local minimum.

$x = \frac{3 π}{2}$ is a local minimum

Step 17.9

These are the local extrema for $f (x) = 2 \sin (x) - \cos^{2} (x)$ .

$x = - \frac{π}{2}$ is a local minimum

$x = \frac{π}{2}$ is a local maximum

$x = \frac{3 π}{2}$ is a local minimum

$x = - \frac{π}{2}$ is a local minimum

$x = \frac{π}{2}$ is a local maximum

$x = \frac{3 π}{2}$ is a local minimum

Step 18