Calculus Examples

$f (x) = 4 x^{3} + 3 x^{2} - 6 x + 1$

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 $4 x^{3} + 3 x^{2} - 6 x + 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$ .

$f' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (1)$

Step 1.1.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 1.1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (1)$

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 $n = 3$ .

$f' (x) = 4 (3 x^{2}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (1)$

Step 1.1.2.3

Multiply $3$ by $4$ .

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

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 $n = 2$ .

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

Step 1.1.3.3

Multiply $2$ by $3$ .

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

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

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

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 $n = 1$ .

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

Step 1.1.4.3

Multiply $- 6$ by $1$ .

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

Step 1.1.5

Differentiate using the Constant Rule.

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Step 1.1.5.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 1.1.5.2

Add $12 x^{2} + 6 x - 6$ and $0$ .

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $12 x^{2} + 6 x - 6$ .

$12 x^{2} + 6 x - 6$

Step 2

Set the first derivative equal to $0$ then solve the equation $12 x^{2} + 6 x - 6 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$12 x^{2} + 6 x - 6 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $6$ out of $12 x^{2} + 6 x - 6$ .

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Step 2.2.1.1

Factor $6$ out of $12 x^{2}$ .

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

Step 2.2.1.2

Factor $6$ out of $6 x$ .

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

Step 2.2.1.3

Factor $6$ out of $- 6$ .

$6 (2 x^{2}) + 6 (x) + 6 (- 1) = 0$

Step 2.2.1.4

Factor $6$ out of $6 (2 x^{2}) + 6 (x)$ .

$6 (2 x^{2} + x) + 6 (- 1) = 0$

Step 2.2.1.5

Factor $6$ out of $6 (2 x^{2} + x) + 6 (- 1)$ .

$6 (2 x^{2} + x - 1) = 0$

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 $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

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Step 2.2.2.1.1.1

Multiply by $1$ .

$6 (2 x^{2} + 1 x - 1) = 0$

Step 2.2.2.1.1.2

Rewrite $1$ as $- 1$ plus $2$

$6 (2 x^{2} + (- 1 + 2) x - 1) = 0$

Step 2.2.2.1.1.3

Apply the distributive property.

$6 (2 x^{2} - 1 x + 2 x - 1) = 0$

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.

$6 ((2 x^{2} - 1 x) + 2 x - 1) = 0$

Step 2.2.2.1.2.2

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

$6 (x (2 x - 1) + 1 (2 x - 1)) = 0$

Step 2.2.2.1.3

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

$6 ((2 x - 1) (x + 1)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$6 (2 x - 1) (x + 1) = 0$

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 $0$ .

$2 x - 1 = 0$

$x + 1 = 0$

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 $x + 1$ 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 $6 (2 x - 1) (x + 1) = 0$ true.

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

Step 3

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

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

Step 4

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

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

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.

$f' (- 2) = 12 {(- 2)}^{2} + 6 (- 2) - 6$

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

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

$f' (- 2) = 12 \cdot 4 + 6 (- 2) - 6$

Step 5.2.1.2

Multiply $12$ by $4$ .

$f' (- 2) = 48 + 6 (- 2) - 6$

Step 5.2.1.3

Multiply $6$ by $- 2$ .

$f' (- 2) = 48 - 12 - 6$

Step 5.2.2

Simplify by subtracting numbers.

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Step 5.2.2.1

Subtract $12$ from $48$ .

$f' (- 2) = 36 - 6$

Step 5.2.2.2

Subtract $6$ from $36$ .

$f' (- 2) = 30$

Step 5.2.3

The final answer is $30$ .

$30$

Step 5.3

At $x = - 2$ the derivative is $30$ . 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 $(- 1, \frac{1}{2})$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.25$ in the expression.

$f' (- 0.25) = 12 {(- 0.25)}^{2} + 6 (- 0.25) - 6$

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

Raise $- 0.25$ to the power of $2$ .

$f' (- 0.25) = 12 \cdot 0.0625 + 6 (- 0.25) - 6$

Step 6.2.1.2

Multiply $12$ by $0.0625$ .

$f' (- 0.25) = 0.75 + 6 (- 0.25) - 6$

Step 6.2.1.3

Multiply $6$ by $- 0.25$ .

$f' (- 0.25) = 0.75 - 1.5 - 6$

Step 6.2.2

Simplify by subtracting numbers.

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Step 6.2.2.1

Subtract $1.5$ from $0.75$ .

$f' (- 0.25) = - 0.75 - 6$

Step 6.2.2.2

Subtract $6$ from $- 0.75$ .

$f' (- 0.25) = - 6.75$

Step 6.2.3

The final answer is $- 6.75$ .

$- 6.75$

Step 6.3

At $x = - 0.25$ the derivative is $- 6.75$ . Since this is negative, the function is decreasing on $(- 1, \frac{1}{2})$ .

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

Step 7

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

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Step 7.1

Replace the variable $x$ with $1.5$ in the expression.

$f' (1.5) = 12 {(1.5)}^{2} + 6 (1.5) - 6$

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 $1.5$ to the power of $2$ .

$f' (1.5) = 12 \cdot 2.25 + 6 (1.5) - 6$

Step 7.2.1.2

Multiply $12$ by $2.25$ .

$f' (1.5) = 27 + 6 (1.5) - 6$

Step 7.2.1.3

Multiply $6$ by $1.5$ .

$f' (1.5) = 27 + 9 - 6$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Add $27$ and $9$ .

$f' (1.5) = 36 - 6$

Step 7.2.2.2

Subtract $6$ from $36$ .

$f' (1.5) = 30$

Step 7.2.3

The final answer is $30$ .

$30$

Step 7.3

At $x = 1.5$ the derivative is $30$ . Since this is positive, the function is increasing on $(\frac{1}{2}, \infty)$ .

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

Step 8

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

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

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

Step 9