Calculus Examples

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

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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

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

Step 1.1.1.2

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

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

Step 1.1.1.3

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

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Step 1.1.1.3.1

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

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

Step 1.1.1.3.2

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

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

Step 1.1.1.3.3

Multiply $- 1$ by $- 1$ .

$\cos (x) + 1 \sin (x)$

Step 1.1.1.3.4

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

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

Step 1.1.3

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

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

Step 1.2

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

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Step 1.2.1

Set the second derivative equal to $0$ .

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

Step 1.2.2

Divide each term in the equation by $\cos (x)$ .

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

Step 1.2.3

Separate fractions.

$\frac{- 1}{1} \cdot \frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 1.2.4

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$\frac{- 1}{1} \cdot \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 1.2.5

Divide $- 1$ by $1$ .

$- \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 1.2.6

Cancel the common factor of $\cos (x)$ .

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Step 1.2.6.1

Cancel the common factor.

$- \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 1.2.6.2

Rewrite the expression.

$- \tan (x) + 1 = \frac{0}{\cos (x)}$

Step 1.2.7

Separate fractions.

$- \tan (x) + 1 = \frac{0}{1} \cdot \frac{1}{\cos (x)}$

Step 1.2.8

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$- \tan (x) + 1 = \frac{0}{1} \cdot \sec (x)$

Step 1.2.9

Divide $0$ by $1$ .

$- \tan (x) + 1 = 0 \sec (x)$

Step 1.2.10

Multiply $0$ by $\sec (x)$ .

$- \tan (x) + 1 = 0$

Step 1.2.11

Subtract $1$ from both sides of the equation.

$- \tan (x) = - 1$

Step 1.2.12

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

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Step 1.2.12.1

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

$\frac{- \tan (x)}{- 1} = \frac{- 1}{- 1}$

Step 1.2.12.2

Simplify the left side.

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Step 1.2.12.2.1

Dividing two negative values results in a positive value.

$\frac{\tan (x)}{1} = \frac{- 1}{- 1}$

Step 1.2.12.2.2

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

$\tan (x) = \frac{- 1}{- 1}$

Step 1.2.12.3

Simplify the right side.

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Step 1.2.12.3.1

Divide $- 1$ by $- 1$ .

$\tan (x) = 1$

Step 1.2.13

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

$x = \arctan (1)$

Step 1.2.14

Simplify the right side.

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Step 1.2.14.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 1.2.15

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + \frac{π}{4}$

Step 1.2.16

Simplify $π + \frac{π}{4}$ .

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Step 1.2.16.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 1.2.16.2

Combine fractions.

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Step 1.2.16.2.1

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 1.2.16.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 1.2.16.3

Simplify the numerator.

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Step 1.2.16.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 1.2.16.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 1.2.17

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

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Step 1.2.17.1

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

$\frac{π}{| b |}$

Step 1.2.17.2

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

$\frac{π}{| 1 |}$

Step 1.2.17.3

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

$\frac{π}{1}$

Step 1.2.17.4

Divide $π$ by $1$ .

$π$

Step 1.2.18

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

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, \infty)$

Step 4

Substitute any number from the interval $(- \infty, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

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

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

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

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

Step 4.2.1.2

Multiply $- 1$ by $0$ .

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

Step 4.2.1.3

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

$f'' (0) = 0 + 1$

Step 4.2.2

Add $0$ and $1$ .

$f'' (0) = 1$

Step 4.2.3

The final answer is $1$ .

$1$

Step 4.3

The graph is concave up on the interval $(- \infty, \infty)$ because $f'' (0)$ is positive.

Concave up on $(- \infty, \infty)$ since $f'' (x)$ is positive

Step 5