Calculus Examples

$f (x) = \frac{1}{3} x^{3} - 3 x^{2} + 9 x + 20$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $\frac{1}{3} x^{3} - 3 x^{2} + 9 x + 20$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$ .

$\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.2

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

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Step 1.1.2.1

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

$\frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.2.2

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

$\frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.2.3

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

$\frac{3}{3} x^{2} + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.2.4

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

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.2.5

Cancel the common factor of $3$ .

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Step 1.1.2.5.1

Cancel the common factor.

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.2.5.2

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

$x^{2} + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 3 x^{2}]$ .

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Step 1.1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{2}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{2}]$ .

$x^{2} - 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.3.2

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

$x^{2} - 3 (2 x) + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.3.3

Multiply $2$ by $- 3$ .

$x^{2} - 6 x + \frac{d}{d x} [9 x] + \frac{d}{d x} [20]$

Step 1.1.4

Evaluate $\frac{d}{d x} [9 x]$ .

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Step 1.1.4.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$x^{2} - 6 x + 9 \frac{d}{d x} [x] + \frac{d}{d x} [20]$

Step 1.1.4.2

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

$x^{2} - 6 x + 9 \cdot 1 + \frac{d}{d x} [20]$

Step 1.1.4.3

Multiply $9$ by $1$ .

$x^{2} - 6 x + 9 + \frac{d}{d x} [20]$

Step 1.1.5

Differentiate using the Constant Rule.

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Step 1.1.5.1

Since $20$ is constant with respect to $x$ , the derivative of $20$ with respect to $x$ is $0$ .

$x^{2} - 6 x + 9 + 0$

Step 1.1.5.2

Add $x^{2} - 6 x + 9$ and $0$ .

$f' (x) = x^{2} - 6 x + 9$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $x^{2} - 6 x + 9$ .

$x^{2} - 6 x + 9$

Step 2

Set the first derivative equal to $0$ then solve the equation $x^{2} - 6 x + 9 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$x^{2} - 6 x + 9 = 0$

Step 2.2

Factor using the perfect square rule.

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Step 2.2.1

Rewrite $9$ as $3^{2}$ .

$x^{2} - 6 x + 3^{2} = 0$

Step 2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 2.2.3

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 3 + 3^{2} = 0$

Step 2.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Step 2.3

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

$x - 3 = 0$

Step 2.4

Add $3$ to both sides of the equation.

$x = 3$

Step 3

The values which make the derivative equal to $0$ are $3$ .

$3$

Step 4

After finding the point that makes the derivative $f' (x) = x^{2} - 6 x + 9$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{1}{3} x^{3} - 3 x^{2} + 9 x + 20$ is increasing and where it is decreasing is $(- \infty, 3) \cup (3, \infty)$ .

$(- \infty, 3) \cup (3, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 3)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $2$ in the expression.

$f' (2) = {(2)}^{2} - 6 \cdot 2 + 9$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

$f' (2) = 4 - 6 \cdot 2 + 9$

Step 5.2.1.2

Multiply $- 6$ by $2$ .

$f' (2) = 4 - 12 + 9$

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Subtract $12$ from $4$ .

$f' (2) = - 8 + 9$

Step 5.2.2.2

Add $- 8$ and $9$ .

$f' (2) = 1$

Step 5.2.3

The final answer is $1$ .

$1$

Step 5.3

At $x = 2$ the derivative is $1$ . Since this is positive, the function is increasing on $(- \infty, 3)$ .

Increasing on $(- \infty, 3)$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(3, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = {(4)}^{2} - 6 \cdot 4 + 9$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Raise $4$ to the power of $2$ .

$f' (4) = 16 - 6 \cdot 4 + 9$

Step 6.2.1.2

Multiply $- 6$ by $4$ .

$f' (4) = 16 - 24 + 9$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Subtract $24$ from $16$ .

$f' (4) = - 8 + 9$

Step 6.2.2.2

Add $- 8$ and $9$ .

$f' (4) = 1$

Step 6.2.3

The final answer is $1$ .

$1$

Step 6.3

At $x = 4$ the derivative is $1$ . Since this is positive, the function is increasing on $(3, \infty)$ .

Increasing on $(3, \infty)$ since $f' (x) > 0$

Step 7

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, 3), (3, \infty)$

Step 8