Calculus Examples

$f (x) = x^{3} - 3 x^{2} - 9 x + 10$

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

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{3} - 3 x^{2} - 9 x + 10$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [10]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (10)$

Step 1.1.1.2

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

$f' (x) = 3 x^{2} + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (10)$

Step 1.1.2

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

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Step 1.1.2.1

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

$f' (x) = 3 x^{2} - 3 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (10)$

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 $n = 2$ .

$f' (x) = 3 x^{2} - 3 (2 x) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (10)$

Step 1.1.2.3

Multiply $2$ by $- 3$ .

$f' (x) = 3 x^{2} - 6 x + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (10)$

Step 1.1.3

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

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Step 1.1.3.1

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

$f' (x) = 3 x^{2} - 6 x - 9 \frac{d}{d x} x + \frac{d}{d x} (10)$

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 $n = 1$ .

$f' (x) = 3 x^{2} - 6 x - 9 \cdot 1 + \frac{d}{d x} (10)$

Step 1.1.3.3

Multiply $- 9$ by $1$ .

$f' (x) = 3 x^{2} - 6 x - 9 + \frac{d}{d x} (10)$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$f' (x) = 3 x^{2} - 6 x - 9 + 0$

Step 1.1.4.2

Add $3 x^{2} - 6 x - 9$ and $0$ .

$f' (x) = 3 x^{2} - 6 x - 9$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} - 6 x - 9$ .

$3 x^{2} - 6 x - 9$

Step 2

Set the first derivative equal to $0$ then solve the equation $3 x^{2} - 6 x - 9 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$3 x^{2} - 6 x - 9 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $3$ out of $3 x^{2} - 6 x - 9$ .

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Step 2.2.1.1

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) - 6 x - 9 = 0$

Step 2.2.1.2

Factor $3$ out of $- 6 x$ .

$3 (x^{2}) + 3 (- 2 x) - 9 = 0$

Step 2.2.1.3

Factor $3$ out of $- 9$ .

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

Step 2.2.1.4

Factor $3$ out of $3 x^{2} + 3 (- 2 x)$ .

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

Step 2.2.1.5

Factor $3$ out of $3 (x^{2} - 2 x) + 3 \cdot - 3$ .

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

Step 2.2.2

Factor.

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Step 2.2.2.1

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

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Step 2.2.2.1.1

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 2.2.2.1.2

Write the factored form using these integers.

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

Step 2.2.2.2

Remove unnecessary parentheses.

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

Step 2.3

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 2.4

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

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Step 2.4.1

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

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

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

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Step 2.5.1

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

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

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

$x = 3, - 1$

Step 3

The values which make the derivative equal to $0$ are $3, - 1$ .

$3, - 1$

Step 4

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 1) \cup (- 1, 3) \cup (3, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 1)$ 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.

$f' (- 2) = 3 {(- 2)}^{2} - 6 \cdot - 2 - 9$

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 $2$ .

$f' (- 2) = 3 \cdot 4 - 6 \cdot - 2 - 9$

Step 5.2.1.2

Multiply $3$ by $4$ .

$f' (- 2) = 12 - 6 \cdot - 2 - 9$

Step 5.2.1.3

Multiply $- 6$ by $- 2$ .

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

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Add $12$ and $12$ .

$f' (- 2) = 24 - 9$

Step 5.2.2.2

Subtract $9$ from $24$ .

$f' (- 2) = 15$

Step 5.2.3

The final answer is $15$ .

$15$

Step 5.3

At $x = - 2$ the derivative is $15$ . Since this is positive, the function is increasing on $(- \infty, - 1)$ .

Increasing on $(- \infty, - 1)$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(- 1, 3)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

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

$f' (1) = 3 {(1)}^{2} - 6 \cdot 1 - 9$

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

One to any power is one.

$f' (1) = 3 \cdot 1 - 6 \cdot 1 - 9$

Step 6.2.1.2

Multiply $3$ by $1$ .

$f' (1) = 3 - 6 \cdot 1 - 9$

Step 6.2.1.3

Multiply $- 6$ by $1$ .

$f' (1) = 3 - 6 - 9$

Step 6.2.2

Simplify by subtracting numbers.

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Step 6.2.2.1

Subtract $6$ from $3$ .

$f' (1) = - 3 - 9$

Step 6.2.2.2

Subtract $9$ from $- 3$ .

$f' (1) = - 12$

Step 6.2.3

The final answer is $- 12$ .

$- 12$

Step 6.3

At $x = 1$ the derivative is $- 12$ . Since this is negative, the function is decreasing on $(- 1, 3)$ .

Decreasing on $(- 1, 3)$ since $f' (x) < 0$

Step 7

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 7.1

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

$f' (4) = 3 {(4)}^{2} - 6 \cdot 4 - 9$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

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

$f' (4) = 3 \cdot 16 - 6 \cdot 4 - 9$

Step 7.2.1.2

Multiply $3$ by $16$ .

$f' (4) = 48 - 6 \cdot 4 - 9$

Step 7.2.1.3

Multiply $- 6$ by $4$ .

$f' (4) = 48 - 24 - 9$

Step 7.2.2

Simplify by subtracting numbers.

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Step 7.2.2.1

Subtract $24$ from $48$ .

$f' (4) = 24 - 9$

Step 7.2.2.2

Subtract $9$ from $24$ .

$f' (4) = 15$

Step 7.2.3

The final answer is $15$ .

$15$

Step 7.3

At $x = 4$ the derivative is $15$ . Since this is positive, the function is increasing on $(3, \infty)$ .

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

Step 8

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

Increasing on: $(- \infty, - 1), (3, \infty)$

Decreasing on: $(- 1, 3)$

Step 9