Calculus Examples

$y = \frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + 7$

Step 1

Write $y = \frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + 7$ as a function.

$f (x) = \frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + 7$

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 $\frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + 7$ with respect to $x$ is $\frac{d}{d x} [\frac{x^{3}}{3}] + \frac{d}{d x} [\frac{x^{2}}{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [7]$ .

$f' (x) = \frac{d}{d x} (\frac{x^{3}}{3}) + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

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 $\frac{1}{3} \frac{d}{d x} [x^{3}]$ .

$f' (x) = \frac{1}{3} \cdot \frac{d}{d x} (x^{3}) + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

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 $n = 3$ .

$f' (x) = \frac{1}{3} \cdot (3 x^{2}) + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

Step 2.1.2.3

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

$f' (x) = \frac{3}{3} x^{2} + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

Step 2.1.2.4

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

$f' (x) = \frac{3 x^{2}}{3} + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

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.

$f' (x) = \frac{3 x^{2}}{3} + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

Step 2.1.2.5.2

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

$f' (x) = x^{2} + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

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 $\frac{1}{2} \frac{d}{d x} [x^{2}]$ .

$f' (x) = x^{2} + \frac{1}{2} \cdot \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

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 $n = 2$ .

$f' (x) = x^{2} + \frac{1}{2} \cdot (2 x) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

Step 2.1.3.3

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

$f' (x) = x^{2} + \frac{2}{2} x + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

Step 2.1.3.4

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

$f' (x) = x^{2} + \frac{2 x}{2} + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

Step 2.1.3.5

Cancel the common factor of $2$ .

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Step 2.1.3.5.1

Cancel the common factor.

$f' (x) = x^{2} + \frac{2 x}{2} + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

Step 2.1.3.5.2

Divide $x$ by $1$ .

$f' (x) = x^{2} + x + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (7)$

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

$f' (x) = x^{2} + x - 2 \frac{d}{d x} x + \frac{d}{d x} (7)$

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 $n = 1$ .

$f' (x) = x^{2} + x - 2 \cdot 1 + \frac{d}{d x} (7)$

Step 2.1.4.3

Multiply $- 2$ by $1$ .

$f' (x) = x^{2} + x - 2 + \frac{d}{d x} (7)$

Step 2.1.5

Differentiate using the Constant Rule.

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Step 2.1.5.1

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$f' (x) = x^{2} + x - 2 + 0$

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

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

$f'' (x) = 2 x + 1 + 0$

Step 2.2.5

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

Subtract $1$ from 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 3.3.3

Simplify the right side.

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Step 3.3.3.1

Move the negative in front of the fraction.

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

Step 4

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

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Step 4.1

Substitute $- \frac{1}{2}$ in $f (x) = \frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + 7$ 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.

$f (- \frac{1}{2}) = \frac{{(- \frac{1}{2})}^{3}}{3} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

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

Simplify the numerator.

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Step 4.1.2.1.1.1

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

$f (- \frac{1}{2}) = \frac{{(- 1)}^{3} {(\frac{1}{2})}^{3}}{3} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.1.2

Raise $- 1$ to the power of $3$ .

$f (- \frac{1}{2}) = \frac{- {(\frac{1}{2})}^{3}}{3} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.1.3

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

$f (- \frac{1}{2}) = \frac{- \frac{1^{3}}{2^{3}}}{3} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.1.4

One to any power is one.

$f (- \frac{1}{2}) = \frac{- \frac{1}{2^{3}}}{3} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.1.5

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

$f (- \frac{1}{2}) = \frac{- \frac{1}{8}}{3} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.2

Multiply the numerator by the reciprocal of the denominator.

$f (- \frac{1}{2}) = - \frac{1}{8} \cdot \frac{1}{3} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.3

Multiply $- \frac{1}{8} \cdot \frac{1}{3}$ .

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Step 4.1.2.1.3.1

Multiply $\frac{1}{3}$ by $\frac{1}{8}$ .

$f (- \frac{1}{2}) = - \frac{1}{3 \cdot 8} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.3.2

Multiply $3$ by $8$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{{(- \frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.4

Simplify the numerator.

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Step 4.1.2.1.4.1

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

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{{(- 1)}^{2} {(\frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.4.2

Raise $- 1$ to the power of $2$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1 {(\frac{1}{2})}^{2}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.4.3

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

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1 (\frac{1^{2}}{2^{2}})}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.4.4

One to any power is one.

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1 (\frac{1}{2^{2}})}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.4.5

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

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1 (\frac{1}{4})}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.4.6

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

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{\frac{1}{4}}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.5

Multiply the numerator by the reciprocal of the denominator.

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{4} \cdot \frac{1}{2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.6

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

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Step 4.1.2.1.6.1

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

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{4 \cdot 2} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.6.2

Multiply $4$ by $2$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{8} - 2 (- \frac{1}{2}) + 7$

Step 4.1.2.1.7

Cancel the common factor of $2$ .

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Step 4.1.2.1.7.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{8} - 2 (\frac{- 1}{2}) + 7$

Step 4.1.2.1.7.2

Factor $2$ out of $- 2$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{8} + 2 (- 1) (\frac{- 1}{2}) + 7$

Step 4.1.2.1.7.3

Cancel the common factor.

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{8} + 2 \cdot (- 1 (\frac{- 1}{2})) + 7$

Step 4.1.2.1.7.4

Rewrite the expression.

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{8} - 1 \cdot - 1 + 7$

Step 4.1.2.1.8

Multiply $- 1$ by $- 1$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{8} + 1 + 7$

Step 4.1.2.2

Find the common denominator.

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Step 4.1.2.2.1

Multiply $\frac{1}{8}$ by $\frac{3}{3}$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{1}{8} \cdot \frac{3}{3} + 1 + 7$

Step 4.1.2.2.2

Multiply $\frac{1}{8}$ by $\frac{3}{3}$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{8 \cdot 3} + 1 + 7$

Step 4.1.2.2.3

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

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{8 \cdot 3} + \frac{1}{1} + 7$

Step 4.1.2.2.4

Multiply $\frac{1}{1}$ by $\frac{24}{24}$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{8 \cdot 3} + \frac{1}{1} \cdot \frac{24}{24} + 7$

Step 4.1.2.2.5

Multiply $\frac{1}{1}$ by $\frac{24}{24}$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{8 \cdot 3} + \frac{24}{24} + 7$

Step 4.1.2.2.6

Write $7$ as a fraction with denominator $1$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{8 \cdot 3} + \frac{24}{24} + \frac{7}{1}$

Step 4.1.2.2.7

Multiply $\frac{7}{1}$ by $\frac{24}{24}$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{8 \cdot 3} + \frac{24}{24} + \frac{7}{1} \cdot \frac{24}{24}$

Step 4.1.2.2.8

Multiply $\frac{7}{1}$ by $\frac{24}{24}$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{8 \cdot 3} + \frac{24}{24} + \frac{7 \cdot 24}{24}$

Step 4.1.2.2.9

Reorder the factors of $8 \cdot 3$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{3 \cdot 8} + \frac{24}{24} + \frac{7 \cdot 24}{24}$

Step 4.1.2.2.10

Multiply $3$ by $8$ .

$f (- \frac{1}{2}) = - \frac{1}{24} + \frac{3}{24} + \frac{24}{24} + \frac{7 \cdot 24}{24}$

Step 4.1.2.3

Combine the numerators over the common denominator.

$f (- \frac{1}{2}) = \frac{- 1 + 3 + 24 + 7 \cdot 24}{24}$

Step 4.1.2.4

Simplify the expression.

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Step 4.1.2.4.1

Multiply $7$ by $24$ .

$f (- \frac{1}{2}) = \frac{- 1 + 3 + 24 + 168}{24}$

Step 4.1.2.4.2

Add $- 1$ and $3$ .

$f (- \frac{1}{2}) = \frac{2 + 24 + 168}{24}$

Step 4.1.2.4.3

Add $2$ and $24$ .

$f (- \frac{1}{2}) = \frac{26 + 168}{24}$

Step 4.1.2.4.4

Add $26$ and $168$ .

$f (- \frac{1}{2}) = \frac{194}{24}$

Step 4.1.2.5

Cancel the common factor of $194$ and $24$ .

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Step 4.1.2.5.1

Factor $2$ out of $194$ .

$f (- \frac{1}{2}) = \frac{2 (97)}{24}$

Step 4.1.2.5.2

Cancel the common factors.

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Step 4.1.2.5.2.1

Factor $2$ out of $24$ .

$f (- \frac{1}{2}) = \frac{2 \cdot 97}{2 \cdot 12}$

Step 4.1.2.5.2.2

Cancel the common factor.

$f (- \frac{1}{2}) = \frac{2 \cdot 97}{2 \cdot 12}$

Step 4.1.2.5.2.3

Rewrite the expression.

$f (- \frac{1}{2}) = \frac{97}{12}$

Step 4.1.2.6

The final answer is $\frac{97}{12}$ .

$\frac{97}{12}$

Step 4.2

The point found by substituting $- \frac{1}{2}$ in $f (x) = \frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + 7$ is $(- \frac{1}{2}, \frac{97}{12})$ . This point can be an inflection point.

$(- \frac{1}{2}, \frac{97}{12})$

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 $- 0.6$ in the expression.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $- 0.6$ .

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

Step 6.2.2

Add $- 1.2$ and $1$ .

$f'' (- 0.6) = - 0.2$

Step 6.2.3

The final answer is $- 0.2$ .

$- 0.2$

Step 6.3

At $- 0.6$ , 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 $- 0.4$ in the expression.

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

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $- 0.4$ .

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

Step 7.2.2

Add $- 0.8$ and $1$ .

$f'' (- 0.4) = 0.2$

Step 7.2.3

The final answer is $0.2$ .

$0.2$

Step 7.3

At $- 0.4$ , 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 $(- \frac{1}{2}, \frac{97}{12})$ .

$(- \frac{1}{2}, \frac{97}{12})$

Step 9