Calculus Examples

$\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x + \frac{1}{3}$

Step 1

Write $\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x + \frac{1}{3}$ as a function.

$f (x) = \frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x + \frac{1}{3}$

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 $\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x + \frac{1}{3}$ with respect to $x$ is $\frac{d}{d x} [\frac{x^{3}}{3}] + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$ .

$\frac{d}{d x} [\frac{x^{3}}{3}] + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.2

Evaluate $\frac{d}{d x} [\frac{x^{3}}{3}]$ .

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Step 2.1.2.1

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{x^{3}}{3}$ with respect to $x$ is $\frac{1}{3} \frac{d}{d x} [x^{3}]$ .

$\frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$\frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.2.3

Combine $3$ and $\frac{1}{3}$ .

$\frac{3}{3} x^{2} + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.2.4

Combine $\frac{3}{3}$ and $x^{2}$ .

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.2.5

Cancel the common factor of $3$ .

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Step 2.1.2.5.1

Cancel the common factor.

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.2.5.2

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{d}{d x} [- \frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3

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

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Step 2.1.3.1

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

$x^{2} - \frac{1}{2} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.2

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

$x^{2} - \frac{1}{2} (2 x) + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.3

Multiply $2$ by $- 1$ .

$x^{2} - 2 (\frac{1}{2}) x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.4

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

$x^{2} + \frac{- 2}{2} x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.5

Combine $\frac{- 2}{2}$ and $x$ .

$x^{2} + \frac{- 2 x}{2} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.6

Cancel the common factor of $- 2$ and $2$ .

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Step 2.1.3.6.1

Factor $2$ out of $- 2 x$ .

$x^{2} + \frac{2 (- x)}{2} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.6.2

Cancel the common factors.

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Step 2.1.3.6.2.1

Factor $2$ out of $2$ .

$x^{2} + \frac{2 (- x)}{2 (1)} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.6.2.2

Cancel the common factor.

$x^{2} + \frac{2 (- x)}{2 \cdot 1} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.6.2.3

Rewrite the expression.

$x^{2} + \frac{- x}{1} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.3.6.2.4

Divide $- x$ by $1$ .

$x^{2} - x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.4

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 2.1.4.1

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

$x^{2} - x - 2 \frac{d}{d x} [x] + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.4.2

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

$x^{2} - x - 2 \cdot 1 + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.4.3

Multiply $- 2$ by $1$ .

$x^{2} - x - 2 + \frac{d}{d x} [\frac{1}{3}]$

Step 2.1.5

Differentiate using the Constant Rule.

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Step 2.1.5.1

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{1}{3}$ with respect to $x$ is $0$ .

$x^{2} - x - 2 + 0$

Step 2.1.5.2

Add $x^{2} - x - 2$ and $0$ .

$f' (x) = x^{2} - x - 2$

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate.

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Step 2.2.1.1

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

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

Step 2.2.1.2

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

$2 x + \frac{d}{d x} [- x] + \frac{d}{d x} [- 2]$

Step 2.2.2

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

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Step 2.2.2.1

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

$2 x - \frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 2.2.2.2

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

$2 x - 1 \cdot 1 + \frac{d}{d x} [- 2]$

Step 2.2.2.3

Multiply $- 1$ by $1$ .

$2 x - 1 + \frac{d}{d x} [- 2]$

Step 2.2.3

Differentiate using the Constant Rule.

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Step 2.2.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$2 x - 1 + 0$

Step 2.2.3.2

Add $2 x - 1$ and $0$ .

$f'' (x) = 2 x - 1$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $2 x - 1$ .

$2 x - 1$

Step 3

Set the second derivative equal to $0$ then solve the equation $2 x - 1 = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$2 x - 1 = 0$

Step 3.2

Add $1$ to both sides of the equation.

$2 x = 1$

Step 3.3

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 3.3.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 3.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 4

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

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Step 4.1

Substitute $\frac{1}{2}$ in $f (x) = \frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x + \frac{1}{3}$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Combine the numerators over the common denominator.

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

Step 4.1.2.2

Simplify each term.

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Step 4.1.2.2.1

Apply the product rule to $\frac{1}{2}$ .

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

Step 4.1.2.2.2

One to any power is one.

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

Step 4.1.2.2.3

Raise $2$ to the power of $3$ .

$f (\frac{1}{2}) = - 2 (\frac{1}{2}) + \frac{\frac{1}{8} + 1}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.3

Simplify the expression.

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Step 4.1.2.3.1

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

$f (\frac{1}{2}) = - 2 (\frac{1}{2}) + \frac{\frac{1}{8} + \frac{8}{8}}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.3.2

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = - 2 (\frac{1}{2}) + \frac{\frac{1 + 8}{8}}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.3.3

Add $1$ and $8$ .

$f (\frac{1}{2}) = - 2 (\frac{1}{2}) + \frac{\frac{9}{8}}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4

Simplify each term.

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Step 4.1.2.4.1

Cancel the common factor of $2$ .

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Step 4.1.2.4.1.1

Factor $2$ out of $- 2$ .

$f (\frac{1}{2}) = 2 (- 1) (\frac{1}{2}) + \frac{\frac{9}{8}}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4.1.2

Cancel the common factor.

$f (\frac{1}{2}) = 2 \cdot (- 1 (\frac{1}{2})) + \frac{\frac{9}{8}}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4.1.3

Rewrite the expression.

$f (\frac{1}{2}) = - 1 + \frac{\frac{9}{8}}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4.2

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{1}{2}) = - 1 + \frac{9}{8} \cdot \frac{1}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4.3

Cancel the common factor of $3$ .

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Step 4.1.2.4.3.1

Factor $3$ out of $9$ .

$f (\frac{1}{2}) = - 1 + \frac{3 (3)}{8} \cdot \frac{1}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4.3.2

Cancel the common factor.

$f (\frac{1}{2}) = - 1 + \frac{3 \cdot 3}{8} \cdot \frac{1}{3} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4.3.3

Rewrite the expression.

$f (\frac{1}{2}) = - 1 + \frac{3}{8} + \frac{- {(\frac{1}{2})}^{2}}{2}$

Step 4.1.2.4.4

Simplify the numerator.

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Step 4.1.2.4.4.1

Apply the product rule to $\frac{1}{2}$ .

$f (\frac{1}{2}) = - 1 + \frac{3}{8} + \frac{- \frac{1^{2}}{2^{2}}}{2}$

Step 4.1.2.4.4.2

One to any power is one.

$f (\frac{1}{2}) = - 1 + \frac{3}{8} + \frac{- \frac{1}{2^{2}}}{2}$

Step 4.1.2.4.4.3

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

$f (\frac{1}{2}) = - 1 + \frac{3}{8} + \frac{- \frac{1}{4}}{2}$

Step 4.1.2.4.5

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{1}{2}) = - 1 + \frac{3}{8} - \frac{1}{4} \cdot \frac{1}{2}$

Step 4.1.2.4.6

Multiply $- \frac{1}{4} \cdot \frac{1}{2}$ .

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Step 4.1.2.4.6.1

Multiply $\frac{1}{2}$ by $\frac{1}{4}$ .

$f (\frac{1}{2}) = - 1 + \frac{3}{8} - \frac{1}{2 \cdot 4}$

Step 4.1.2.4.6.2

Multiply $2$ by $4$ .

$f (\frac{1}{2}) = - 1 + \frac{3}{8} - \frac{1}{8}$

Step 4.1.2.5

Simplify terms.

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Step 4.1.2.5.1

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = - 1 + \frac{3 - 1}{8}$

Step 4.1.2.5.2

Subtract $1$ from $3$ .

$f (\frac{1}{2}) = - 1 + \frac{2}{8}$

Step 4.1.2.5.3

Cancel the common factor of $2$ and $8$ .

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Step 4.1.2.5.3.1

Factor $2$ out of $2$ .

$f (\frac{1}{2}) = - 1 + \frac{2 (1)}{8}$

Step 4.1.2.5.3.2

Cancel the common factors.

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Step 4.1.2.5.3.2.1

Factor $2$ out of $8$ .

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

Step 4.1.2.5.3.2.2

Cancel the common factor.

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

Step 4.1.2.5.3.2.3

Rewrite the expression.

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

Step 4.1.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 4.1.2.7

Combine $- 1$ and $\frac{4}{4}$ .

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

Step 4.1.2.8

Combine the numerators over the common denominator.

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

Step 4.1.2.9

Simplify the numerator.

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Step 4.1.2.9.1

Multiply $- 1$ by $4$ .

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

Step 4.1.2.9.2

Add $- 4$ and $1$ .

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

Step 4.1.2.10

Move the negative in front of the fraction.

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

Step 4.1.2.11

The final answer is $- \frac{3}{4}$ .

$- \frac{3}{4}$

Step 4.2

The point found by substituting $\frac{1}{2}$ in $f (x) = \frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x + \frac{1}{3}$ is $(\frac{1}{2}, - \frac{3}{4})$ . This point can be an inflection point.

$(\frac{1}{2}, - \frac{3}{4})$

Step 5

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

$(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.4$ in the expression.

$f'' (0.4) = 2 (0.4) - 1$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $0.4$ .

$f'' (0.4) = 0.8 - 1$

Step 6.2.2

Subtract $1$ from $0.8$ .

$f'' (0.4) = - 0.2$

Step 6.2.3

The final answer is $- 0.2$ .

$- 0.2$

Step 6.3

At $0.4$ , the second derivative is $- 0.2$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{1}{2})$

Decreasing on $(- \infty, \frac{1}{2})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{1}{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.6$ in the expression.

$f'' (0.6) = 2 (0.6) - 1$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $0.6$ .

$f'' (0.6) = 1.2 - 1$

Step 7.2.2

Subtract $1$ from $1.2$ .

$f'' (0.6) = 0.2$

Step 7.2.3

The final answer is $0.2$ .

$0.2$

Step 7.3

At $0.6$ , the second derivative is $0.2$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{2}, \infty)$ .

Increasing on $(\frac{1}{2}, \infty)$ 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 $(\frac{1}{2}, - \frac{3}{4})$ .

$(\frac{1}{2}, - \frac{3}{4})$

Step 9