Calculus Examples

$f (x) = 4 - 12 x + 6 x^{2} - x^{3}$

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 $4 - 12 x + 6 x^{2} - x^{3}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- 12 x] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- x^{3}]$ .

$f' (x) = \frac{d}{d x} (4) + \frac{d}{d x} (- 12 x) + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (- x^{3})$

Step 1.1.1.2

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

$f' (x) = 0 + \frac{d}{d x} (- 12 x) + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (- x^{3})$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 12 x]$ .

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Step 1.1.2.1

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

$f' (x) = 0 - 12 \frac{d}{d x} x + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (- x^{3})$

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 $n = 1$ .

$f' (x) = 0 - 12 \cdot 1 + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (- x^{3})$

Step 1.1.2.3

Multiply $- 12$ by $1$ .

$f' (x) = 0 - 12 + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (- x^{3})$

Step 1.1.3

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

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Step 1.1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 0 - 12 + 6 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- x^{3})$

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 $n = 2$ .

$f' (x) = 0 - 12 + 6 (2 x) + \frac{d}{d x} (- x^{3})$

Step 1.1.3.3

Multiply $2$ by $6$ .

$f' (x) = 0 - 12 + 12 x + \frac{d}{d x} (- x^{3})$

Step 1.1.4

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

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Step 1.1.4.1

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

$f' (x) = 0 - 12 + 12 x - \frac{d}{d x} x^{3}$

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 $n = 3$ .

$f' (x) = 0 - 12 + 12 x - (3 x^{2})$

Step 1.1.4.3

Multiply $3$ by $- 1$ .

$f' (x) = 0 - 12 + 12 x - 3 x^{2}$

Step 1.1.5

Simplify.

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Step 1.1.5.1

Subtract $12$ from $0$ .

$f' (x) = - 12 + 12 x - 3 x^{2}$

Step 1.1.5.2

Reorder terms.

$f' (x) = - 3 x^{2} + 12 x - 12$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 3 x^{2} + 12 x - 12$ .

$- 3 x^{2} + 12 x - 12$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 3 x^{2} + 12 x - 12 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- 3 x^{2} + 12 x - 12 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

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

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Step 2.2.1.1

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

$- 3 x^{2} + 12 x - 12 = 0$

Step 2.2.1.2

Factor $- 3$ out of $12 x$ .

$- 3 x^{2} - 3 (- 4 x) - 12 = 0$

Step 2.2.1.3

Factor $- 3$ out of $- 12$ .

$- 3 x^{2} - 3 (- 4 x) - 3 \cdot 4 = 0$

Step 2.2.1.4

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

$- 3 (x^{2} - 4 x) - 3 \cdot 4 = 0$

Step 2.2.1.5

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

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

Step 2.2.2

Factor using the perfect square rule.

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Step 2.2.2.1

Rewrite $4$ as $2^{2}$ .

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

Step 2.2.2.2

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

$4 x = 2 \cdot x \cdot 2$

Step 2.2.2.3

Rewrite the polynomial.

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

Step 2.2.2.4

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

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

Step 2.3

Divide each term in $- 3 {(x - 2)}^{2} = 0$ by $- 3$ and simplify.

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Step 2.3.1

Divide each term in $- 3 {(x - 2)}^{2} = 0$ by $- 3$ .

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $- 3$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

Divide ${(x - 2)}^{2}$ by $1$ .

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0$ by $- 3$ .

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

Step 2.4

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

$x - 2 = 0$

Step 2.5

Add $2$ to both sides of the equation.

$x = 2$

Step 3

The values which make the derivative equal to $0$ are $2$ .

$2$

Step 4

After finding the point that makes the derivative $f' (x) = - 3 x^{2} + 12 x - 12$ equal to $0$ or undefined, the interval to check where $f (x) = 4 - 12 x + 6 x^{2} - x^{3}$ is increasing and where it is decreasing is $(- \infty, 2) \cup (2, \infty)$ .

$(- \infty, 2) \cup (2, \infty)$

Step 5

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

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Step 5.1

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

$f' (1) = - 3 {(1)}^{2} + 12 (1) - 12$

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

One to any power is one.

$f' (1) = - 3 \cdot 1 + 12 (1) - 12$

Step 5.2.1.2

Multiply $- 3$ by $1$ .

$f' (1) = - 3 + 12 (1) - 12$

Step 5.2.1.3

Multiply $12$ by $1$ .

$f' (1) = - 3 + 12 - 12$

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Add $- 3$ and $12$ .

$f' (1) = 9 - 12$

Step 5.2.2.2

Subtract $12$ from $9$ .

$f' (1) = - 3$

Step 5.2.3

The final answer is $- 3$ .

$- 3$

Step 5.3

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

Decreasing on $(- \infty, 2)$ since $f' (x) < 0$

Step 6

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

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Step 6.1

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

$f' (3) = - 3 {(3)}^{2} + 12 (3) - 12$

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 $3$ to the power of $2$ .

$f' (3) = - 3 \cdot 9 + 12 (3) - 12$

Step 6.2.1.2

Multiply $- 3$ by $9$ .

$f' (3) = - 27 + 12 (3) - 12$

Step 6.2.1.3

Multiply $12$ by $3$ .

$f' (3) = - 27 + 36 - 12$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Add $- 27$ and $36$ .

$f' (3) = 9 - 12$

Step 6.2.2.2

Subtract $12$ from $9$ .

$f' (3) = - 3$

Step 6.2.3

The final answer is $- 3$ .

$- 3$

Step 6.3

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

Decreasing on $(2, \infty)$ since $f' (x) < 0$

Step 7

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

Decreasing on: $(- \infty, 2), (2, \infty)$

Step 8