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 second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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 2.1.2

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

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Step 2.1.2.1

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 2.1.2.2

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

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

Step 2.1.3

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

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Step 2.1.3.1

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 2.1.3.2

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 2.1.3.2.1

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 2.1.3.2.2

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 2.1.3.2.3

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

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

Step 2.1.3.3

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

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

Step 2.1.3.4

Multiply $- 1$ by $2$ .

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

Step 2.1.3.5

Multiply $- 2$ by $- 1$ .

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

Step 2.1.4

Simplify.

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Step 2.1.4.1

Reorder terms.

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

Step 2.1.4.2

Simplify each term.

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Step 2.1.4.2.1

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

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

Step 2.1.4.2.2

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

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

Step 2.1.4.2.3

Apply the sine double-angle identity.

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

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

Step 2.2.2

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

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Step 2.2.2.1

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 2.2.2.1.1

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

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

Step 2.2.2.1.2

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

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

Step 2.2.2.1.3

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

$\cos (2 x) (2 \cdot 1) + \frac{d}{d x} [2 \cos (x)]$

Step 2.2.2.4

Multiply $2$ by $1$ .

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

Step 2.2.2.5

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

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

Step 2.2.3

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

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

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

Step 2.2.3.3

Multiply $- 1$ by $2$ .

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

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $2 \cos (2 x) - 2 \sin (x)$ .

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

Step 3

Set the second derivative equal to $0$ then solve the equation $2 \cos (2 x) - 2 \sin (x) = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

Simplify each term.

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Step 3.2.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

$2 (1 - 2 \sin^{2} (x)) - 2 \sin (x) = 0$

Step 3.2.2

Apply the distributive property.

$2 \cdot 1 + 2 (- 2 \sin^{2} (x)) - 2 \sin (x) = 0$

Step 3.2.3

Multiply $2$ by $1$ .

$2 + 2 (- 2 \sin^{2} (x)) - 2 \sin (x) = 0$

Step 3.2.4

Multiply $- 2$ by $2$ .

$2 - 4 \sin^{2} (x) - 2 \sin (x) = 0$

Step 3.3

Factor $2 - 4 \sin^{2} (x) - 2 \sin (x)$ .

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Step 3.3.1

Factor $2$ out of $2 - 4 \sin^{2} (x) - 2 \sin (x)$ .

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Step 3.3.1.1

Factor $2$ out of $2$ .

$2 (1) - 4 \sin^{2} (x) - 2 \sin (x) = 0$

Step 3.3.1.2

Factor $2$ out of $- 4 \sin^{2} (x)$ .

$2 (1) + 2 (- 2 \sin^{2} (x)) - 2 \sin (x) = 0$

Step 3.3.1.3

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

$2 (1) + 2 (- 2 \sin^{2} (x)) + 2 (- \sin (x)) = 0$

Step 3.3.1.4

Factor $2$ out of $2 (1) + 2 (- 2 \sin^{2} (x))$ .

$2 (1 - 2 \sin^{2} (x)) + 2 (- \sin (x)) = 0$

Step 3.3.1.5

Factor $2$ out of $2 (1 - 2 \sin^{2} (x)) + 2 (- \sin (x))$ .

$2 (1 - 2 \sin^{2} (x) - \sin (x)) = 0$

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor by grouping.

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Step 3.3.2.1.1

Reorder terms.

$2 (- 2 \sin^{2} (x) - \sin (x) + 1) = 0$

Step 3.3.2.1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 2 \cdot 1 = - 2$ and whose sum is $b = - 1$ .

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Step 3.3.2.1.2.1

Factor $- 1$ out of $- \sin (x)$ .

$2 (- 2 \sin^{2} (x) - \sin (x) + 1) = 0$

Step 3.3.2.1.2.2

Rewrite $- 1$ as $1$ plus $- 2$

$2 (- 2 \sin^{2} (x) + (1 - 2) \sin (x) + 1) = 0$

Step 3.3.2.1.2.3

Apply the distributive property.

$2 (- 2 \sin^{2} (x) + 1 \sin (x) - 2 \sin (x) + 1) = 0$

Step 3.3.2.1.2.4

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

$2 (- 2 \sin^{2} (x) + \sin (x) - 2 \sin (x) + 1) = 0$

Step 3.3.2.1.3

Factor out the greatest common factor from each group.

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Step 3.3.2.1.3.1

Group the first two terms and the last two terms.

$2 (- 2 \sin^{2} (x) + \sin (x) - 2 \sin (x) + 1) = 0$

Step 3.3.2.1.3.2

Factor out the greatest common factor (GCF) from each group.

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

Step 3.3.2.1.4

Factor the polynomial by factoring out the greatest common factor, $- 2 \sin (x) + 1$ .

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

Step 3.3.2.2

Remove unnecessary parentheses.

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

Step 3.4

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

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

$\sin (x) + 1 = 0$

Step 3.5

Set $- 2 \sin (x) + 1$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $- 2 \sin (x) + 1$ equal to $0$ .

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

Step 3.5.2

Solve $- 2 \sin (x) + 1 = 0$ for $x$ .

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Step 3.5.2.1

Subtract $1$ from both sides of the equation.

$- 2 \sin (x) = - 1$

Step 3.5.2.2

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ and simplify.

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Step 3.5.2.2.1

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ .

$\frac{- 2 \sin (x)}{- 2} = \frac{- 1}{- 2}$

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $- 2$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{- 2 \sin (x)}{- 2} = \frac{- 1}{- 2}$

Step 3.5.2.2.2.1.2

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

$\sin (x) = \frac{- 1}{- 2}$

Step 3.5.2.2.3

Simplify the right side.

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Step 3.5.2.2.3.1

Dividing two negative values results in a positive value.

$\sin (x) = \frac{1}{2}$

Step 3.5.2.3

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

$x = \arcsin (\frac{1}{2})$

Step 3.5.2.4

Simplify the right side.

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Step 3.5.2.4.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 3.5.2.5

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.

$x = π - \frac{π}{6}$

Step 3.5.2.6

Simplify $π - \frac{π}{6}$ .

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Step 3.5.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 3.5.2.6.2

Combine fractions.

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Step 3.5.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 3.5.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 3.5.2.6.3

Simplify the numerator.

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Step 3.5.2.6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 3.5.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 3.5.2.7

Find the period of $\sin (x)$ .

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Step 3.5.2.7.1

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

$\frac{2 π}{| b |}$

Step 3.5.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.5.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 3.5.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.5.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 3.6

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

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Step 3.6.1

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

$\sin (x) + 1 = 0$

Step 3.6.2

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

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Step 3.6.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 3.6.2.2

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

$x = \arcsin (- 1)$

Step 3.6.2.3

Simplify the right side.

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Step 3.6.2.3.1

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

$x = - \frac{π}{2}$

Step 3.6.2.4

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 3.6.2.5

Simplify the expression to find the second solution.

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Step 3.6.2.5.1

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

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 3.6.2.5.2

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 3.6.2.6

Find the period of $\sin (x)$ .

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Step 3.6.2.6.1

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

$\frac{2 π}{| b |}$

Step 3.6.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.6.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 3.6.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.6.2.7

Add $2 π$ to every negative angle to get positive angles.

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Step 3.6.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 3.6.2.7.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 3.6.2.7.3

Combine fractions.

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Step 3.6.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 3.6.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 3.6.2.7.4

Simplify the numerator.

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Step 3.6.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 3.6.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 3.6.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 3.6.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 3.7

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

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 3.8

Consolidate the answers.

$x = \frac{π}{6} + \frac{2 π n}{3}$ , for any integer $n$

Step 4

Find the points where the second derivative is $0$ .

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Step 4.1

Substitute $\frac{π}{6}$ in $f (x) = 2 \sin (x) - \cos^{2} (x)$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = 2 \sin (\frac{π}{6}) - \cos^{2} (\frac{π}{6})$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$f (\frac{π}{6}) = 2 (\frac{1}{2}) - \cos^{2} (\frac{π}{6})$

Step 4.1.2.1.2

Cancel the common factor of $2$ .

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Step 4.1.2.1.2.1

Cancel the common factor.

$f (\frac{π}{6}) = 2 (\frac{1}{2}) - \cos^{2} (\frac{π}{6})$

Step 4.1.2.1.2.2

Rewrite the expression.

$f (\frac{π}{6}) = 1 - \cos^{2} (\frac{π}{6})$

Step 4.1.2.1.3

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = 1 - {(\frac{\sqrt{3}}{2})}^{2}$

Step 4.1.2.1.4

Apply the product rule to $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = 1 - \frac{{\sqrt{3}}^{2}}{2^{2}}$

Step 4.1.2.1.5

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 4.1.2.1.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (\frac{π}{6}) = 1 - \frac{{(3^{\frac{1}{2}})}^{2}}{2^{2}}$

Step 4.1.2.1.5.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{π}{6}) = 1 - \frac{3^{\frac{1}{2} \cdot 2}}{2^{2}}$

Step 4.1.2.1.5.3

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{π}{6}) = 1 - \frac{3^{\frac{2}{2}}}{2^{2}}$

Step 4.1.2.1.5.4

Cancel the common factor of $2$ .

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Step 4.1.2.1.5.4.1

Cancel the common factor.

$f (\frac{π}{6}) = 1 - \frac{3^{\frac{2}{2}}}{2^{2}}$

Step 4.1.2.1.5.4.2

Rewrite the expression.

$f (\frac{π}{6}) = 1 - \frac{3}{2^{2}}$

Step 4.1.2.1.5.5

Evaluate the exponent.

$f (\frac{π}{6}) = 1 - \frac{3}{2^{2}}$

Step 4.1.2.1.6

Raise $2$ to the power of $2$ .

$f (\frac{π}{6}) = 1 - \frac{3}{4}$

Step 4.1.2.2

Simplify the expression.

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Step 4.1.2.2.1

Write $1$ as a fraction with a common denominator.

$f (\frac{π}{6}) = \frac{4}{4} - \frac{3}{4}$

Step 4.1.2.2.2

Combine the numerators over the common denominator.

$f (\frac{π}{6}) = \frac{4 - 3}{4}$

Step 4.1.2.2.3

Subtract $3$ from $4$ .

$f (\frac{π}{6}) = \frac{1}{4}$

Step 4.1.2.3

The final answer is $\frac{1}{4}$ .

$\frac{1}{4}$

Step 4.2

The point found by substituting $\frac{π}{6}$ in $f (x) = 2 \sin (x) - \cos^{2} (x)$ is $(\frac{π}{6}, \frac{1}{4})$ . This point can be an inflection point.

$(\frac{π}{6}, \frac{1}{4})$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, \frac{π}{6}) \cup (\frac{π}{6}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{π}{6})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.42359877$ in the expression.

$f'' (0.42359877) = 2 \cos (2 (0.42359877)) - 2 \sin (0.42359877)$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $0.42359877$ .

$f'' (0.42359877) = 2 \cos (0.84719755) - 2 \sin (0.42359877)$

Step 6.2.2

The final answer is $2 \cos (0.84719755) - 2 \sin (0.42359877)$ .

$2 \cos (0.84719755) - 2 \sin (0.42359877)$

Step 6.3

At $0.42359877$ , the second derivative is $0.50208433$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, \frac{π}{6})$ .

Increasing on $(- \infty, \frac{π}{6})$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(\frac{π}{6}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.62359877$ in the expression.

$f'' (0.62359877) = 2 \cos (2 (0.62359877)) - 2 \sin (0.62359877)$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $0.62359877$ .

$f'' (0.62359877) = 2 \cos (1.24719755) - 2 \sin (0.62359877)$

Step 7.2.2

The final answer is $2 \cos (1.24719755) - 2 \sin (0.62359877)$ .

$2 \cos (1.24719755) - 2 \sin (0.62359877)$

Step 7.3

At $0.62359877$ , the second derivative is $- 0.53195951$ . Since this is negative, the second derivative is decreasing on the interval $(\frac{π}{6}, \infty)$

Decreasing on $(\frac{π}{6}, \infty)$ since $f'' (x) < 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(\frac{π}{6}, \frac{1}{4})$ .

$(\frac{π}{6}, \frac{1}{4})$

Step 9