Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $\frac{1}{3} x^{3} - x^{2} - 3 x + 4$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.2

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

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Step 1.2.1

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

$\frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.2.2

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

$\frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.2.3

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

$\frac{3}{3} x^{2} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.2.4

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

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.2.5

Cancel the common factor of $3$ .

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Step 1.2.5.1

Cancel the common factor.

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.2.5.2

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

$x^{2} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.3

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

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Step 1.3.1

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

$x^{2} - \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.3.2

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

$x^{2} - (2 x) + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.3.3

Multiply $2$ by $- 1$ .

$x^{2} - 2 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 1.4

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

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Step 1.4.1

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

$x^{2} - 2 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 1.4.2

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

$x^{2} - 2 x - 3 \cdot 1 + \frac{d}{d x} [4]$

Step 1.4.3

Multiply $- 3$ by $1$ .

$x^{2} - 2 x - 3 + \frac{d}{d x} [4]$

Step 1.5

Differentiate using the Constant Rule.

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Step 1.5.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$x^{2} - 2 x - 3 + 0$

Step 1.5.2

Add $x^{2} - 2 x - 3$ and $0$ .

$x^{2} - 2 x - 3$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x^{2} - 2 x - 3$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [- 3]$ .

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $- 2$ by $1$ .

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

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

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

Step 2.3.2

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

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$x^{2} - 2 x - 3 = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

$\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.2

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

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Step 4.1.2.1

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

$\frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.2.2

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

$\frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.2.3

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

$\frac{3}{3} x^{2} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.2.4

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

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.2.5

Cancel the common factor of $3$ .

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Step 4.1.2.5.1

Cancel the common factor.

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.2.5.2

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

$x^{2} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.3

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

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Step 4.1.3.1

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

$x^{2} - \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.3.2

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

$x^{2} - (2 x) + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.3.3

Multiply $2$ by $- 1$ .

$x^{2} - 2 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

Step 4.1.4

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

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Step 4.1.4.1

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

$x^{2} - 2 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 4.1.4.2

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

$x^{2} - 2 x - 3 \cdot 1 + \frac{d}{d x} [4]$

Step 4.1.4.3

Multiply $- 3$ by $1$ .

$x^{2} - 2 x - 3 + \frac{d}{d x} [4]$

Step 4.1.5

Differentiate using the Constant Rule.

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Step 4.1.5.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$x^{2} - 2 x - 3 + 0$

Step 4.1.5.2

Add $x^{2} - 2 x - 3$ and $0$ .

$f' (x) = x^{2} - 2 x - 3$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $x^{2} - 2 x - 3$ .

$x^{2} - 2 x - 3$

Step 5

Set the first derivative equal to $0$ then solve the equation $x^{2} - 2 x - 3 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$x^{2} - 2 x - 3 = 0$

Step 5.2

Factor $x^{2} - 2 x - 3$ using the AC method.

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Step 5.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 5.2.2

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

Step 5.3

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

$x - 3 = 0$

$x + 1 = 0$

Step 5.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 5.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 5.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 5.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 5.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 5.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.6

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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.

Step 7

Critical points to evaluate.

$x = 3, - 1$

Step 8

Evaluate the second derivative at $x = 3$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 (3) - 2$

Step 9

Evaluate the second derivative.

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Step 9.1

Multiply $2$ by $3$ .

$6 - 2$

Step 9.2

Subtract $2$ from $6$ .

$4$

Step 10

$x = 3$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 3$ is a local minimum

Step 11

Find the y-value when $x = 3$ .

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Step 11.1

Replace the variable $x$ with $3$ in the expression.

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

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Cancel the common factor of $3$ .

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Step 11.2.1.1.1

Factor $3$ out of ${(3)}^{3}$ .

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

Step 11.2.1.1.2

Cancel the common factor.

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

Step 11.2.1.1.3

Rewrite the expression.

$f (3) = 3^{2} - {(3)}^{2} - 3 \cdot 3 + 4$

Step 11.2.1.2

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

$f (3) = 9 - {(3)}^{2} - 3 \cdot 3 + 4$

Step 11.2.1.3

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

$f (3) = 9 - 1 \cdot 9 - 3 \cdot 3 + 4$

Step 11.2.1.4

Multiply $- 1$ by $9$ .

$f (3) = 9 - 9 - 3 \cdot 3 + 4$

Step 11.2.1.5

Multiply $- 3$ by $3$ .

$f (3) = 9 - 9 - 9 + 4$

Step 11.2.2

Simplify by adding and subtracting.

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Step 11.2.2.1

Subtract $9$ from $9$ .

$f (3) = 0 - 9 + 4$

Step 11.2.2.2

Subtract $9$ from $0$ .

$f (3) = - 9 + 4$

Step 11.2.2.3

Add $- 9$ and $4$ .

$f (3) = - 5$

Step 11.2.3

The final answer is $- 5$ .

$y = - 5$

Step 12

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 (- 1) - 2$

Step 13

Evaluate the second derivative.

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Step 13.1

Multiply $2$ by $- 1$ .

$- 2 - 2$

Step 13.2

Subtract $2$ from $- 2$ .

$- 4$

Step 14

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

Step 15

Find the y-value when $x = - 1$ .

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Step 15.1

Replace the variable $x$ with $- 1$ in the expression.

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

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

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

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

Step 15.2.1.2

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

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

Step 15.2.1.3

Move the negative in front of the fraction.

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

Step 15.2.1.4

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 15.2.1.4.1

Multiply $- 1$ by ${(- 1)}^{2}$ .

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Step 15.2.1.4.1.1

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

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

Step 15.2.1.4.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 15.2.1.4.2

Add $1$ and $2$ .

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

Step 15.2.1.5

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

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

Step 15.2.1.6

Multiply $- 3$ by $- 1$ .

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

Step 15.2.2

Find the common denominator.

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Step 15.2.2.1

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

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

Step 15.2.2.2

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

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

Step 15.2.2.3

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

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

Step 15.2.2.4

Write $3$ as a fraction with denominator $1$ .

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

Step 15.2.2.5

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

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

Step 15.2.2.6

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

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

Step 15.2.2.7

Write $4$ as a fraction with denominator $1$ .

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

Step 15.2.2.8

Multiply $\frac{4}{1}$ by $\frac{3}{3}$ .

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

Step 15.2.2.9

Multiply $\frac{4}{1}$ by $\frac{3}{3}$ .

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

Step 15.2.3

Combine the numerators over the common denominator.

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

Step 15.2.4

Simplify each term.

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Step 15.2.4.1

Multiply $- 1$ by $3$ .

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

Step 15.2.4.2

Multiply $3$ by $3$ .

$f (- 1) = \frac{- 1 - 3 + 9 + 4 \cdot 3}{3}$

Step 15.2.4.3

Multiply $4$ by $3$ .

$f (- 1) = \frac{- 1 - 3 + 9 + 12}{3}$

Step 15.2.5

Simplify by adding and subtracting.

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Step 15.2.5.1

Subtract $3$ from $- 1$ .

$f (- 1) = \frac{- 4 + 9 + 12}{3}$

Step 15.2.5.2

Add $- 4$ and $9$ .

$f (- 1) = \frac{5 + 12}{3}$

Step 15.2.5.3

Add $5$ and $12$ .

$f (- 1) = \frac{17}{3}$

Step 15.2.6

The final answer is $\frac{17}{3}$ .

$y = \frac{17}{3}$

Step 16

These are the local extrema for $f (x) = \frac{1}{3} x^{3} - x^{2} - 3 x + 4$ .

$(3, - 5)$ is a local minima

$(- 1, \frac{17}{3})$ is a local maxima

Step 17