Calculus Examples

$4 x^{3} + 3 x^{2} - 6 x + 1$

Step 1

Write $4 x^{3} + 3 x^{2} - 6 x + 1$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $4 x^{3} + 3 x^{2} - 6 x + 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 2.2

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

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Step 2.2.1

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

$4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$4 (3 x^{2}) + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 2.2.3

Multiply $3$ by $4$ .

$12 x^{2} + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 2.3

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

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Step 2.3.1

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

$12 x^{2} + 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 2.3.2

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

$12 x^{2} + 3 (2 x) + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 2.3.3

Multiply $2$ by $3$ .

$12 x^{2} + 6 x + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 2.4

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

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Step 2.4.1

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

$12 x^{2} + 6 x - 6 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 2.4.2

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

$12 x^{2} + 6 x - 6 \cdot 1 + \frac{d}{d x} [1]$

Step 2.4.3

Multiply $- 6$ by $1$ .

$12 x^{2} + 6 x - 6 + \frac{d}{d x} [1]$

Step 2.5

Differentiate using the Constant Rule.

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Step 2.5.1

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

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

Step 2.5.2

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

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

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $12 x^{2} + 6 x - 6$ with respect to $x$ is $\frac{d}{d x} [12 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 6]$ .

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

Step 3.2

Evaluate $\frac{d}{d x} [12 x^{2}]$ .

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Step 3.2.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{2}$ with respect to $x$ is $12 \frac{d}{d x} [x^{2}]$ .

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

Step 3.2.2

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

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

Step 3.2.3

Multiply $2$ by $12$ .

$f'' (x) = 24 x + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 6)$

Step 3.3

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

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Step 3.3.1

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

$f'' (x) = 24 x + 6 \frac{d}{d x} (x) + \frac{d}{d x} (- 6)$

Step 3.3.2

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

$f'' (x) = 24 x + 6 \cdot 1 + \frac{d}{d x} (- 6)$

Step 3.3.3

Multiply $6$ by $1$ .

$f'' (x) = 24 x + 6 + \frac{d}{d x} (- 6)$

Step 3.4

Differentiate using the Constant Rule.

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Step 3.4.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$f'' (x) = 24 x + 6 + 0$

Step 3.4.2

Add $24 x + 6$ and $0$ .

$f'' (x) = 24 x + 6$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

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

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

By the Sum Rule, the derivative of $4 x^{3} + 3 x^{2} - 6 x + 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 5.1.2

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

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Step 5.1.2.1

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

$4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 5.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$4 (3 x^{2}) + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 5.1.2.3

Multiply $3$ by $4$ .

$12 x^{2} + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 5.1.3

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

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Step 5.1.3.1

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

$12 x^{2} + 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 5.1.3.2

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

$12 x^{2} + 3 (2 x) + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 5.1.3.3

Multiply $2$ by $3$ .

$12 x^{2} + 6 x + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [1]$

Step 5.1.4

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

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Step 5.1.4.1

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

$12 x^{2} + 6 x - 6 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 5.1.4.2

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

$12 x^{2} + 6 x - 6 \cdot 1 + \frac{d}{d x} [1]$

Step 5.1.4.3

Multiply $- 6$ by $1$ .

$12 x^{2} + 6 x - 6 + \frac{d}{d x} [1]$

Step 5.1.5

Differentiate using the Constant Rule.

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Step 5.1.5.1

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

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

Step 5.1.5.2

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

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

Step 5.2

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

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

Step 6

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

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Step 6.1

Set the first derivative equal to $0$ .

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

Step 6.2

Factor the left side of the equation.

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Step 6.2.1

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

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Step 6.2.1.1

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

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

Step 6.2.1.2

Factor $6$ out of $6 x$ .

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

Step 6.2.1.3

Factor $6$ out of $- 6$ .

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

Step 6.2.1.4

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

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

Step 6.2.1.5

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

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

Step 6.2.2

Factor.

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Step 6.2.2.1

Factor by grouping.

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Step 6.2.2.1.1

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 6.2.2.1.1.1

Multiply by $1$ .

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

Step 6.2.2.1.1.2

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

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

Step 6.2.2.1.1.3

Apply the distributive property.

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

Step 6.2.2.1.2

Factor out the greatest common factor from each group.

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Step 6.2.2.1.2.1

Group the first two terms and the last two terms.

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

Step 6.2.2.1.2.2

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

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

Step 6.2.2.1.3

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

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

Step 6.2.2.2

Remove unnecessary parentheses.

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

Step 6.3

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 6.4

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

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Step 6.4.1

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

$2 x - 1 = 0$

Step 6.4.2

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

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Step 6.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 6.4.2.2

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

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Step 6.4.2.2.1

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

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

Step 6.4.2.2.2

Simplify the left side.

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Step 6.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.4.2.2.2.1.1

Cancel the common factor.

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

Step 6.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 6.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 6.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 6.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6.6

The final solution is all the values that make $6 (2 x - 1) (x + 1) = 0$ true.

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

Step 7

Find the values where the derivative is undefined.

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Step 7.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 8

Critical points to evaluate.

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

Step 9

Evaluate the second derivative at $x = \frac{1}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$24 (\frac{1}{2}) + 6$

Step 10

Evaluate the second derivative.

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Step 10.1

Cancel the common factor of $2$ .

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Step 10.1.1

Factor $2$ out of $24$ .

$2 (12) \frac{1}{2} + 6$

Step 10.1.2

Cancel the common factor.

$2 \cdot 12 \frac{1}{2} + 6$

Step 10.1.3

Rewrite the expression.

$12 + 6$

Step 10.2

Add $12$ and $6$ .

$18$

Step 11

$x = \frac{1}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{1}{2}$ is a local minimum

Step 12

Find the y-value when $x = \frac{1}{2}$ .

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Step 12.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

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

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

Apply the product rule to $\frac{1}{2}$ .

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

Step 12.2.1.2

One to any power is one.

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

Step 12.2.1.3

Raise $2$ to the power of $3$ .

$f (\frac{1}{2}) = 4 (\frac{1}{8}) + 3 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 1$

Step 12.2.1.4

Cancel the common factor of $4$ .

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Step 12.2.1.4.1

Factor $4$ out of $8$ .

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

Step 12.2.1.4.2

Cancel the common factor.

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

Step 12.2.1.4.3

Rewrite the expression.

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

Step 12.2.1.5

Apply the product rule to $\frac{1}{2}$ .

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

Step 12.2.1.6

One to any power is one.

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

Step 12.2.1.7

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

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

Step 12.2.1.8

Combine $3$ and $\frac{1}{4}$ .

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

Step 12.2.1.9

Cancel the common factor of $2$ .

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Step 12.2.1.9.1

Factor $2$ out of $- 6$ .

$f (\frac{1}{2}) = \frac{1}{2} + \frac{3}{4} + 2 (- 3) (\frac{1}{2}) + 1$

Step 12.2.1.9.2

Cancel the common factor.

$f (\frac{1}{2}) = \frac{1}{2} + \frac{3}{4} + 2 \cdot (- 3 (\frac{1}{2})) + 1$

Step 12.2.1.9.3

Rewrite the expression.

$f (\frac{1}{2}) = \frac{1}{2} + \frac{3}{4} - 3 + 1$

Step 12.2.2

Find the common denominator.

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Step 12.2.2.1

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{1}{2} \cdot \frac{2}{2} + \frac{3}{4} - 3 + 1$

Step 12.2.2.2

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{2}{2 \cdot 2} + \frac{3}{4} - 3 + 1$

Step 12.2.2.3

Write $- 3$ as a fraction with denominator $1$ .

$f (\frac{1}{2}) = \frac{2}{2 \cdot 2} + \frac{3}{4} + \frac{- 3}{1} + 1$

Step 12.2.2.4

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{2}{2 \cdot 2} + \frac{3}{4} + \frac{- 3}{1} \cdot \frac{4}{4} + 1$

Step 12.2.2.5

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{2}{2 \cdot 2} + \frac{3}{4} + \frac{- 3 \cdot 4}{4} + 1$

Step 12.2.2.6

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

$f (\frac{1}{2}) = \frac{2}{2 \cdot 2} + \frac{3}{4} + \frac{- 3 \cdot 4}{4} + \frac{1}{1}$

Step 12.2.2.7

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

$f (\frac{1}{2}) = \frac{2}{2 \cdot 2} + \frac{3}{4} + \frac{- 3 \cdot 4}{4} + \frac{1}{1} \cdot \frac{4}{4}$

Step 12.2.2.8

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

$f (\frac{1}{2}) = \frac{2}{2 \cdot 2} + \frac{3}{4} + \frac{- 3 \cdot 4}{4} + \frac{4}{4}$

Step 12.2.2.9

Multiply $2$ by $2$ .

$f (\frac{1}{2}) = \frac{2}{4} + \frac{3}{4} + \frac{- 3 \cdot 4}{4} + \frac{4}{4}$

Step 12.2.3

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{2 + 3 - 3 \cdot 4 + 4}{4}$

Step 12.2.4

Simplify the expression.

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Step 12.2.4.1

Multiply $- 3$ by $4$ .

$f (\frac{1}{2}) = \frac{2 + 3 - 12 + 4}{4}$

Step 12.2.4.2

Add $2$ and $3$ .

$f (\frac{1}{2}) = \frac{5 - 12 + 4}{4}$

Step 12.2.4.3

Subtract $12$ from $5$ .

$f (\frac{1}{2}) = \frac{- 7 + 4}{4}$

Step 12.2.4.4

Add $- 7$ and $4$ .

$f (\frac{1}{2}) = \frac{- 3}{4}$

Step 12.2.4.5

Move the negative in front of the fraction.

$f (\frac{1}{2}) = - \frac{3}{4}$

Step 12.2.5

The final answer is $- \frac{3}{4}$ .

$y = - \frac{3}{4}$

Step 13

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$24 (- 1) + 6$

Step 14

Evaluate the second derivative.

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Step 14.1

Multiply $24$ by $- 1$ .

$- 24 + 6$

Step 14.2

Add $- 24$ and $6$ .

$- 18$

Step 15

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

Step 16

Find the y-value when $x = - 1$ .

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Step 16.1

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

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

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify each term.

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Step 16.2.1.1

Raise $- 1$ to the power of $3$ .

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

Step 16.2.1.2

Multiply $4$ by $- 1$ .

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

Step 16.2.1.3

Raise $- 1$ to the power of $2$ .

$f (- 1) = - 4 + 3 \cdot 1 - 6 \cdot - 1 + 1$

Step 16.2.1.4

Multiply $3$ by $1$ .

$f (- 1) = - 4 + 3 - 6 \cdot - 1 + 1$

Step 16.2.1.5

Multiply $- 6$ by $- 1$ .

$f (- 1) = - 4 + 3 + 6 + 1$

Step 16.2.2

Simplify by adding numbers.

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Step 16.2.2.1

Add $- 4$ and $3$ .

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

Step 16.2.2.2

Add $- 1$ and $6$ .

$f (- 1) = 5 + 1$

Step 16.2.2.3

Add $5$ and $1$ .

$f (- 1) = 6$

Step 16.2.3

The final answer is $6$ .

$y = 6$

Step 17

These are the local extrema for $f (x) = 4 x^{3} + 3 x^{2} - 6 x + 1$ .

$(\frac{1}{2}, - \frac{3}{4})$ is a local minima

$(- 1, 6)$ is a local maxima

Step 18