Calculus Examples

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

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

Rewrite ${(3 x - 7)}^{2}$ as $(3 x - 7) (3 x - 7)$ .

$\frac{d}{d x} [(3 x - 7) (3 x - 7)]$

Step 1.1.2

Expand $(3 x - 7) (3 x - 7)$ using the FOIL Method.

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Step 1.1.2.1

Apply the distributive property.

$\frac{d}{d x} [3 x (3 x - 7) - 7 (3 x - 7)]$

Step 1.1.2.2

Apply the distributive property.

$\frac{d}{d x} [3 x (3 x) + 3 x \cdot - 7 - 7 (3 x - 7)]$

Step 1.1.2.3

Apply the distributive property.

$\frac{d}{d x} [3 x (3 x) + 3 x \cdot - 7 - 7 (3 x) - 7 \cdot - 7]$

Step 1.1.3

Simplify and combine like terms.

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Step 1.1.3.1

Simplify each term.

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Step 1.1.3.1.1

Rewrite using the commutative property of multiplication.

$\frac{d}{d x} [3 \cdot 3 x \cdot x + 3 x \cdot - 7 - 7 (3 x) - 7 \cdot - 7]$

Step 1.1.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.3.1.2.1

Move $x$ .

$\frac{d}{d x} [3 \cdot 3 (x \cdot x) + 3 x \cdot - 7 - 7 (3 x) - 7 \cdot - 7]$

Step 1.1.3.1.2.2

Multiply $x$ by $x$ .

$\frac{d}{d x} [3 \cdot 3 x^{2} + 3 x \cdot - 7 - 7 (3 x) - 7 \cdot - 7]$

Step 1.1.3.1.3

Multiply $3$ by $3$ .

$\frac{d}{d x} [9 x^{2} + 3 x \cdot - 7 - 7 (3 x) - 7 \cdot - 7]$

Step 1.1.3.1.4

Multiply $- 7$ by $3$ .

$\frac{d}{d x} [9 x^{2} - 21 x - 7 (3 x) - 7 \cdot - 7]$

Step 1.1.3.1.5

Multiply $3$ by $- 7$ .

$\frac{d}{d x} [9 x^{2} - 21 x - 21 x - 7 \cdot - 7]$

Step 1.1.3.1.6

Multiply $- 7$ by $- 7$ .

$\frac{d}{d x} [9 x^{2} - 21 x - 21 x + 49]$

Step 1.1.3.2

Subtract $21 x$ from $- 21 x$ .

$\frac{d}{d x} [9 x^{2} - 42 x + 49]$

Step 1.1.4

By the Sum Rule, the derivative of $9 x^{2} - 42 x + 49$ with respect to $x$ is $\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- 42 x] + \frac{d}{d x} [49]$ .

$\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- 42 x] + \frac{d}{d x} [49]$

Step 1.1.5

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$9 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 42 x] + \frac{d}{d x} [49]$

Step 1.1.6

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

$9 (2 x) + \frac{d}{d x} [- 42 x] + \frac{d}{d x} [49]$

Step 1.1.7

Multiply $2$ by $9$ .

$18 x + \frac{d}{d x} [- 42 x] + \frac{d}{d x} [49]$

Step 1.1.8

Since $- 42$ is constant with respect to $x$ , the derivative of $- 42 x$ with respect to $x$ is $- 42 \frac{d}{d x} [x]$ .

$18 x - 42 \frac{d}{d x} [x] + \frac{d}{d x} [49]$

Step 1.1.9

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

$18 x - 42 \cdot 1 + \frac{d}{d x} [49]$

Step 1.1.10

Multiply $- 42$ by $1$ .

$18 x - 42 + \frac{d}{d x} [49]$

Step 1.1.11

Since $49$ is constant with respect to $x$ , the derivative of $49$ with respect to $x$ is $0$ .

$18 x - 42 + 0$

Step 1.1.12

Add $18 x - 42$ and $0$ .

$f' (x) = 18 x - 42$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $18 x - 42$ .

$18 x - 42$

Step 2

Set the first derivative equal to $0$ then solve the equation $18 x - 42 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$18 x - 42 = 0$

Step 2.2

Add $42$ to both sides of the equation.

$18 x = 42$

Step 2.3

Divide each term in $18 x = 42$ by $18$ and simplify.

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Step 2.3.1

Divide each term in $18 x = 42$ by $18$ .

$\frac{18 x}{18} = \frac{42}{18}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $18$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{18 x}{18} = \frac{42}{18}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{42}{18}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $42$ and $18$ .

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Step 2.3.3.1.1

Factor $6$ out of $42$ .

$x = \frac{6 (7)}{18}$

Step 2.3.3.1.2

Cancel the common factors.

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Step 2.3.3.1.2.1

Factor $6$ out of $18$ .

$x = \frac{6 \cdot 7}{6 \cdot 3}$

Step 2.3.3.1.2.2

Cancel the common factor.

$x = \frac{6 \cdot 7}{6 \cdot 3}$

Step 2.3.3.1.2.3

Rewrite the expression.

$x = \frac{7}{3}$

Step 3

The values which make the derivative equal to $0$ are $\frac{7}{3}$ .

$\frac{7}{3}$

Step 4

After finding the point that makes the derivative $f' (x) = 18 x - 42$ equal to $0$ or undefined, the interval to check where $f (x) = {(3 x - 7)}^{2}$ is increasing and where it is decreasing is $(- \infty, \frac{7}{3}) \cup (\frac{7}{3}, \infty)$ .

$(- \infty, \frac{7}{3}) \cup (\frac{7}{3}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{7}{3})$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $\frac{4}{3}$ in the expression.

$f' (\frac{4}{3}) = 18 (\frac{4}{3}) - 42$

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

Cancel the common factor of $3$ .

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Step 5.2.1.1.1

Factor $3$ out of $18$ .

$f' (\frac{4}{3}) = 3 (6) (\frac{4}{3}) - 42$

Step 5.2.1.1.2

Cancel the common factor.

$f' (\frac{4}{3}) = 3 \cdot (6 (\frac{4}{3})) - 42$

Step 5.2.1.1.3

Rewrite the expression.

$f' (\frac{4}{3}) = 6 \cdot 4 - 42$

Step 5.2.1.2

Multiply $6$ by $4$ .

$f' (\frac{4}{3}) = 24 - 42$

Step 5.2.2

Subtract $42$ from $24$ .

$f' (\frac{4}{3}) = - 18$

Step 5.2.3

The final answer is $- 18$ .

$- 18$

Step 5.3

At $x = \frac{4}{3}$ the derivative is $- 18$ . Since this is negative, the function is decreasing on $(- \infty, \frac{7}{3})$ .

Decreasing on $(- \infty, \frac{7}{3})$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(\frac{7}{3}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $\frac{10}{3}$ in the expression.

$f' (\frac{10}{3}) = 18 (\frac{10}{3}) - 42$

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

Cancel the common factor of $3$ .

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Step 6.2.1.1.1

Factor $3$ out of $18$ .

$f' (\frac{10}{3}) = 3 (6) (\frac{10}{3}) - 42$

Step 6.2.1.1.2

Cancel the common factor.

$f' (\frac{10}{3}) = 3 \cdot (6 (\frac{10}{3})) - 42$

Step 6.2.1.1.3

Rewrite the expression.

$f' (\frac{10}{3}) = 6 \cdot 10 - 42$

Step 6.2.1.2

Multiply $6$ by $10$ .

$f' (\frac{10}{3}) = 60 - 42$

Step 6.2.2

Subtract $42$ from $60$ .

$f' (\frac{10}{3}) = 18$

Step 6.2.3

The final answer is $18$ .

$18$

Step 6.3

At $x = \frac{10}{3}$ the derivative is $18$ . Since this is positive, the function is increasing on $(\frac{7}{3}, \infty)$ .

Increasing on $(\frac{7}{3}, \infty)$ since $f' (x) > 0$

Step 7

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

Increasing on: $(\frac{7}{3}, \infty)$

Decreasing on: $(- \infty, \frac{7}{3})$

Step 8