Calculus Examples

$y = x - \sin (x)$

Step 1

Write $y = x - \sin (x)$ as a function.

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

Step 2

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

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

Differentiate.

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Step 2.1.1.1.1

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

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

Step 2.1.1.1.2

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

$1 + \frac{d}{d x} [- \sin (x)]$

Step 2.1.1.2

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

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Step 2.1.1.2.1

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

$1 - \frac{d}{d x} [\sin (x)]$

Step 2.1.1.2.2

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

$f' (x) = 1 - \cos (x)$

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

Differentiate.

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Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

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

$0 + \frac{d}{d x} [- \cos (x)]$

Step 2.1.2.2

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

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Step 2.1.2.2.1

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

$0 - \frac{d}{d x} [\cos (x)]$

Step 2.1.2.2.2

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

$0 - - \sin (x)$

Step 2.1.2.2.3

Multiply $- 1$ by $- 1$ .

$0 + 1 \sin (x)$

Step 2.1.2.2.4

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

$0 + \sin (x)$

Step 2.1.2.3

Add $0$ and $\sin (x)$ .

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

Step 2.1.3

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

$\sin (x)$

Step 2.2

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

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Step 2.2.1

Set the second derivative equal to $0$ .

$\sin (x) = 0$

Step 2.2.2

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

$x = \arcsin (0)$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

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

$x = 0$

Step 2.2.4

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.

$x = π - 0$

Step 2.2.5

Subtract $0$ from $π$ .

$x = π$

Step 2.2.6

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

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Step 2.2.6.1

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

$\frac{2 π}{| b |}$

Step 2.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 2.2.6.3

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

$\frac{2 π}{1}$

Step 2.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.2.7

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

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 2.2.8

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 3

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 4

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

$(- \infty, \infty)$

Step 5

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

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Step 5.1

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

$f'' (0) = \sin (0)$

Step 5.2

Simplify the result.

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Step 5.2.1

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

$f'' (0) = 0$

Step 5.2.2

The final answer is $0$ .

$0$

Step 5.3

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 6