Calculus Examples

$f (x) = \frac{1}{3} x^{3} - \frac{5}{2} x^{2} + 6 x - 8$

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 $\frac{1}{3} x^{3} - \frac{5}{2} x^{2} + 6 x - 8$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$ .

$\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

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

$\frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

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 $n = 3$ .

$\frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.2.3

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

$\frac{3}{3} x^{2} + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.2.4

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

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

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.

$\frac{3 x^{2}}{3} + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.2.5.2

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

$x^{2} + \frac{d}{d x} [- \frac{5}{2} x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- \frac{5}{2} x^{2}]$ .

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Step 1.1.3.1

Since $- \frac{5}{2}$ is constant with respect to $x$ , the derivative of $- \frac{5}{2} x^{2}$ with respect to $x$ is $- \frac{5}{2} \frac{d}{d x} [x^{2}]$ .

$x^{2} - \frac{5}{2} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

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 $n = 2$ .

$x^{2} - \frac{5}{2} (2 x) + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.3

Multiply $2$ by $- 1$ .

$x^{2} - 2 (\frac{5}{2}) x + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.4

Combine $- 2$ and $\frac{5}{2}$ .

$x^{2} + \frac{- 2 \cdot 5}{2} x + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.5

Multiply $- 2$ by $5$ .

$x^{2} + \frac{- 10}{2} x + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.6

Combine $\frac{- 10}{2}$ and $x$ .

$x^{2} + \frac{- 10 x}{2} + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.7

Cancel the common factor of $- 10$ and $2$ .

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Step 1.1.3.7.1

Factor $2$ out of $- 10 x$ .

$x^{2} + \frac{2 (- 5 x)}{2} + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.7.2

Cancel the common factors.

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Step 1.1.3.7.2.1

Factor $2$ out of $2$ .

$x^{2} + \frac{2 (- 5 x)}{2 (1)} + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.7.2.2

Cancel the common factor.

$x^{2} + \frac{2 (- 5 x)}{2 \cdot 1} + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.7.2.3

Rewrite the expression.

$x^{2} + \frac{- 5 x}{1} + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.3.7.2.4

Divide $- 5 x$ by $1$ .

$x^{2} - 5 x + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 8]$

Step 1.1.4

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

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Step 1.1.4.1

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

$x^{2} - 5 x + 6 \frac{d}{d x} [x] + \frac{d}{d x} [- 8]$

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 $n = 1$ .

$x^{2} - 5 x + 6 \cdot 1 + \frac{d}{d x} [- 8]$

Step 1.1.4.3

Multiply $6$ by $1$ .

$x^{2} - 5 x + 6 + \frac{d}{d x} [- 8]$

Step 1.1.5

Differentiate using the Constant Rule.

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Step 1.1.5.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$x^{2} - 5 x + 6 + 0$

Step 1.1.5.2

Add $x^{2} - 5 x + 6$ and $0$ .

$f' (x) = x^{2} - 5 x + 6$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $x^{2} - 5 x + 6$ .

$x^{2} - 5 x + 6$

Step 2

Set the first derivative equal to $0$ then solve the equation $x^{2} - 5 x + 6 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$x^{2} - 5 x + 6 = 0$

Step 2.2

Factor $x^{2} - 5 x + 6$ using the AC method.

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Step 2.2.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 2.2.2

Write the factored form using these integers.

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

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 $0$ .

$x - 3 = 0$

$x - 2 = 0$

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

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Step 2.5.1

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

$x - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

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

$x = 3, 2$

Step 3

The values which make the derivative equal to $0$ are $3, 2$ .

$3, 2$

Step 4

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

$(- \infty, 2) \cup (2, 3) \cup (3, \infty)$

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.

$f' (1) = {(1)}^{2} - 5 \cdot 1 + 6$

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.

$f' (1) = 1 - 5 \cdot 1 + 6$

Step 5.2.1.2

Multiply $- 5$ by $1$ .

$f' (1) = 1 - 5 + 6$

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Subtract $5$ from $1$ .

$f' (1) = - 4 + 6$

Step 5.2.2.2

Add $- 4$ and $6$ .

$f' (1) = 2$

Step 5.2.3

The final answer is $2$ .

$2$

Step 5.3

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

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $\frac{5}{2}$ in the expression.

$f' (\frac{5}{2}) = {(\frac{5}{2})}^{2} - 5 (\frac{5}{2}) + 6$

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

Apply the product rule to $\frac{5}{2}$ .

$f' (\frac{5}{2}) = \frac{5^{2}}{2^{2}} - 5 (\frac{5}{2}) + 6$

Step 6.2.1.2

Raise $5$ to the power of $2$ .

$f' (\frac{5}{2}) = \frac{25}{2^{2}} - 5 (\frac{5}{2}) + 6$

Step 6.2.1.3

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

$f' (\frac{5}{2}) = \frac{25}{4} - 5 (\frac{5}{2}) + 6$

Step 6.2.1.4

Multiply $- 5 (\frac{5}{2})$ .

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Step 6.2.1.4.1

Combine $- 5$ and $\frac{5}{2}$ .

$f' (\frac{5}{2}) = \frac{25}{4} + \frac{- 5 \cdot 5}{2} + 6$

Step 6.2.1.4.2

Multiply $- 5$ by $5$ .

$f' (\frac{5}{2}) = \frac{25}{4} + \frac{- 25}{2} + 6$

Step 6.2.1.5

Move the negative in front of the fraction.

$f' (\frac{5}{2}) = \frac{25}{4} - \frac{25}{2} + 6$

Step 6.2.2

Find the common denominator.

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Step 6.2.2.1

Multiply $\frac{25}{2}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{25}{4} - (\frac{25}{2} \cdot \frac{2}{2}) + 6$

Step 6.2.2.2

Multiply $\frac{25}{2}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + 6$

Step 6.2.2.3

Write $6$ as a fraction with denominator $1$ .

$f' (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{6}{1}$

Step 6.2.2.4

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$f' (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{6}{1} \cdot \frac{4}{4}$

Step 6.2.2.5

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$f' (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{6 \cdot 4}{4}$

Step 6.2.2.6

Multiply $2$ by $2$ .

$f' (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{4} + \frac{6 \cdot 4}{4}$

Step 6.2.3

Combine the numerators over the common denominator.

$f' (\frac{5}{2}) = \frac{25 - 25 \cdot 2 + 6 \cdot 4}{4}$

Step 6.2.4

Simplify each term.

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Step 6.2.4.1

Multiply $- 25$ by $2$ .

$f' (\frac{5}{2}) = \frac{25 - 50 + 6 \cdot 4}{4}$

Step 6.2.4.2

Multiply $6$ by $4$ .

$f' (\frac{5}{2}) = \frac{25 - 50 + 24}{4}$

Step 6.2.5

Simplify the expression.

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Step 6.2.5.1

Subtract $50$ from $25$ .

$f' (\frac{5}{2}) = \frac{- 25 + 24}{4}$

Step 6.2.5.2

Add $- 25$ and $24$ .

$f' (\frac{5}{2}) = \frac{- 1}{4}$

Step 6.2.5.3

Move the negative in front of the fraction.

$f' (\frac{5}{2}) = - \frac{1}{4}$

Step 6.2.6

The final answer is $- \frac{1}{4}$ .

$- \frac{1}{4}$

Step 6.3

At $x = \frac{5}{2}$ the derivative is $- \frac{1}{4}$ . Since this is negative, the function is decreasing on $(2, 3)$ .

Decreasing on $(2, 3)$ 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.

$f' (4) = {(4)}^{2} - 5 \cdot 4 + 6$

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 $2$ .

$f' (4) = 16 - 5 \cdot 4 + 6$

Step 7.2.1.2

Multiply $- 5$ by $4$ .

$f' (4) = 16 - 20 + 6$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Subtract $20$ from $16$ .

$f' (4) = - 4 + 6$

Step 7.2.2.2

Add $- 4$ and $6$ .

$f' (4) = 2$

Step 7.2.3

The final answer is $2$ .

$2$

Step 7.3

At $x = 4$ the derivative is $2$ . 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: $(- \infty, 2), (3, \infty)$

Decreasing on: $(2, 3)$

Step 9