Calculus Examples

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

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

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

Step 1.1.2

Expand $(2 x - 3) (2 x - 3)$ using the FOIL Method.

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Step 1.1.2.1

Apply the distributive property.

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

Step 1.1.2.2

Apply the distributive property.

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

Step 1.1.2.3

Apply the distributive property.

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

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.

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

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 $x$ .

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

Step 1.1.3.1.2.2

Multiply $x$ by $x$ .

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

Step 1.1.3.1.3

Multiply $2$ by $2$ .

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

Step 1.1.3.1.4

Multiply $- 3$ by $2$ .

$\frac{d}{d x} [x (4 x^{2} - 6 x - 3 (2 x) - 3 \cdot - 3)]$

Step 1.1.3.1.5

Multiply $2$ by $- 3$ .

$\frac{d}{d x} [x (4 x^{2} - 6 x - 6 x - 3 \cdot - 3)]$

Step 1.1.3.1.6

Multiply $- 3$ by $- 3$ .

$\frac{d}{d x} [x (4 x^{2} - 6 x - 6 x + 9)]$

Step 1.1.3.2

Subtract $6 x$ from $- 6 x$ .

$\frac{d}{d x} [x (4 x^{2} - 12 x + 9)]$

Step 1.1.4

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = 4 x^{2} - 12 x + 9$ .

$x \frac{d}{d x} [4 x^{2} - 12 x + 9] + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5

Differentiate.

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Step 1.1.5.1

By the Sum Rule, the derivative of $4 x^{2} - 12 x + 9$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 12 x] + \frac{d}{d x} [9]$ .

$x (\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 12 x] + \frac{d}{d x} [9]) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$x (4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 12 x] + \frac{d}{d x} [9]) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.3

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

$x (4 (2 x) + \frac{d}{d x} [- 12 x] + \frac{d}{d x} [9]) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.4

Multiply $2$ by $4$ .

$x (8 x + \frac{d}{d x} [- 12 x] + \frac{d}{d x} [9]) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.5

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x$ with respect to $x$ is $- 12 \frac{d}{d x} [x]$ .

$x (8 x - 12 \frac{d}{d x} [x] + \frac{d}{d x} [9]) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.6

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

$x (8 x - 12 \cdot 1 + \frac{d}{d x} [9]) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.7

Multiply $- 12$ by $1$ .

$x (8 x - 12 + \frac{d}{d x} [9]) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.8

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$x (8 x - 12 + 0) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.9

Add $8 x - 12$ and $0$ .

$x (8 x - 12) + (4 x^{2} - 12 x + 9) \frac{d}{d x} [x]$

Step 1.1.5.10

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

$x (8 x - 12) + (4 x^{2} - 12 x + 9) \cdot 1$

Step 1.1.5.11

Multiply $4 x^{2} - 12 x + 9$ by $1$ .

$x (8 x - 12) + 4 x^{2} - 12 x + 9$

Step 1.1.6

Simplify.

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Step 1.1.6.1

Apply the distributive property.

$x (8 x) + x \cdot - 12 + 4 x^{2} - 12 x + 9$

Step 1.1.6.2

Combine terms.

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Step 1.1.6.2.1

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

$8 (x^{1} x) + x \cdot - 12 + 4 x^{2} - 12 x + 9$

Step 1.1.6.2.2

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

$8 (x^{1} x^{1}) + x \cdot - 12 + 4 x^{2} - 12 x + 9$

Step 1.1.6.2.3

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

$8 x^{1 + 1} + x \cdot - 12 + 4 x^{2} - 12 x + 9$

Step 1.1.6.2.4

Add $1$ and $1$ .

$8 x^{2} + x \cdot - 12 + 4 x^{2} - 12 x + 9$

Step 1.1.6.2.5

Move $- 12$ to the left of $x$ .

$8 x^{2} - 12 \cdot x + 4 x^{2} - 12 x + 9$

Step 1.1.6.2.6

Add $8 x^{2}$ and $4 x^{2}$ .

$12 x^{2} - 12 x - 12 x + 9$

Step 1.1.6.2.7

Subtract $12 x$ from $- 12 x$ .

$f' (x) = 12 x^{2} - 24 x + 9$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $12 x^{2} - 24 x + 9$ .

$12 x^{2} - 24 x + 9$

Step 2

Set the first derivative equal to $0$ then solve the equation $12 x^{2} - 24 x + 9 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$12 x^{2} - 24 x + 9 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $3$ out of $12 x^{2} - 24 x + 9$ .

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Step 2.2.1.1

Factor $3$ out of $12 x^{2}$ .

$3 (4 x^{2}) - 24 x + 9 = 0$

Step 2.2.1.2

Factor $3$ out of $- 24 x$ .

$3 (4 x^{2}) + 3 (- 8 x) + 9 = 0$

Step 2.2.1.3

Factor $3$ out of $9$ .

$3 (4 x^{2}) + 3 (- 8 x) + 3 (3) = 0$

Step 2.2.1.4

Factor $3$ out of $3 (4 x^{2}) + 3 (- 8 x)$ .

$3 (4 x^{2} - 8 x) + 3 (3) = 0$

Step 2.2.1.5

Factor $3$ out of $3 (4 x^{2} - 8 x) + 3 (3)$ .

$3 (4 x^{2} - 8 x + 3) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor by grouping.

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Step 2.2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 3 = 12$ and whose sum is $b = - 8$ .

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Step 2.2.2.1.1.1

Factor $- 8$ out of $- 8 x$ .

$3 (4 x^{2} - 8 x + 3) = 0$

Step 2.2.2.1.1.2

Rewrite $- 8$ as $- 2$ plus $- 6$

$3 (4 x^{2} + (- 2 - 6) x + 3) = 0$

Step 2.2.2.1.1.3

Apply the distributive property.

$3 (4 x^{2} - 2 x - 6 x + 3) = 0$

Step 2.2.2.1.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1.2.1

Group the first two terms and the last two terms.

$3 ((4 x^{2} - 2 x) - 6 x + 3) = 0$

Step 2.2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$3 (2 x (2 x - 1) - 3 (2 x - 1)) = 0$

Step 2.2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$3 ((2 x - 1) (2 x - 3)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$3 (2 x - 1) (2 x - 3) = 0$

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 $0$ .

$2 x - 1 = 0$

$2 x - 3 = 0$

Step 2.4

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 2.4.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.4.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 2.4.2.2.1

Divide each term in $2 x = 1$ by $2$ .

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5

Set $2 x - 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Step 2.5.2

Solve $2 x - 3 = 0$ for $x$ .

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Step 2.5.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 2.5.2.2

Divide each term in $2 x = 3$ by $2$ and simplify.

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Step 2.5.2.2.1

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 2.6

The final solution is all the values that make $3 (2 x - 1) (2 x - 3) = 0$ true.

$x = \frac{1}{2}, \frac{3}{2}$

Step 3

The values which make the derivative equal to $0$ are $\frac{1}{2}, \frac{3}{2}$ .

$\frac{1}{2}, \frac{3}{2}$

Step 4

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

$(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{1}{2})$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 0.5$ in the expression.

$f' (- 0.5) = 12 {(- 0.5)}^{2} - 24 \cdot - 0.5 + 9$

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 $- 0.5$ to the power of $2$ .

$f' (- 0.5) = 12 \cdot 0.25 - 24 \cdot - 0.5 + 9$

Step 5.2.1.2

Multiply $12$ by $0.25$ .

$f' (- 0.5) = 3 - 24 \cdot - 0.5 + 9$

Step 5.2.1.3

Multiply $- 24$ by $- 0.5$ .

$f' (- 0.5) = 3 + 12 + 9$

Step 5.2.2

Simplify by adding numbers.

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Step 5.2.2.1

Add $3$ and $12$ .

$f' (- 0.5) = 15 + 9$

Step 5.2.2.2

Add $15$ and $9$ .

$f' (- 0.5) = 24$

Step 5.2.3

The final answer is $24$ .

$24$

Step 5.3

At $x = - 0.5$ the derivative is $24$ . Since this is positive, the function is increasing on $(- \infty, \frac{1}{2})$ .

Increasing on $(- \infty, \frac{1}{2})$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(0.5, \frac{3}{2})$ 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.

$f' (1) = 12 {(1)}^{2} - 24 \cdot 1 + 9$

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.

$f' (1) = 12 \cdot 1 - 24 \cdot 1 + 9$

Step 6.2.1.2

Multiply $12$ by $1$ .

$f' (1) = 12 - 24 \cdot 1 + 9$

Step 6.2.1.3

Multiply $- 24$ by $1$ .

$f' (1) = 12 - 24 + 9$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Subtract $24$ from $12$ .

$f' (1) = - 12 + 9$

Step 6.2.2.2

Add $- 12$ and $9$ .

$f' (1) = - 3$

Step 6.2.3

The final answer is $- 3$ .

$- 3$

Step 6.3

At $x = 1$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(0.5, \frac{3}{2})$ .

Decreasing on $(\frac{1}{2}, \frac{3}{2})$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{3}{2}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $2.5$ in the expression.

$f' (2.5) = 12 {(2.5)}^{2} - 24 \cdot 2.5 + 9$

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 $2.5$ to the power of $2$ .

$f' (2.5) = 12 \cdot 6.25 - 24 \cdot 2.5 + 9$

Step 7.2.1.2

Multiply $12$ by $6.25$ .

$f' (2.5) = 75 - 24 \cdot 2.5 + 9$

Step 7.2.1.3

Multiply $- 24$ by $2.5$ .

$f' (2.5) = 75 - 60 + 9$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Subtract $60$ from $75$ .

$f' (2.5) = 15 + 9$

Step 7.2.2.2

Add $15$ and $9$ .

$f' (2.5) = 24$

Step 7.2.3

The final answer is $24$ .

$24$

Step 7.3

At $x = 2.5$ the derivative is $24$ . Since this is positive, the function is increasing on $(\frac{3}{2}, \infty)$ .

Increasing on $(\frac{3}{2}, \infty)$ since $f' (x) > 0$

Step 8

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

Increasing on: $(- \infty, \frac{1}{2}), (\frac{3}{2}, \infty)$

Decreasing on: $(\frac{1}{2}, \frac{3}{2})$

Step 9