Calculus Examples

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

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate.

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Step 1.1.1.1.1

By the Sum Rule, the derivative of $x^{3} - 3 x^{2} + 12$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [12]$ .

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

Step 1.1.1.1.2

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

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

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

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

Step 1.1.1.2.2

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

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

Step 1.1.1.2.3

Multiply $2$ by $- 3$ .

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

Step 1.1.1.3

Differentiate using the Constant Rule.

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Step 1.1.1.3.1

Since $12$ is constant with respect to $x$ , the derivative of $12$ with respect to $x$ is $0$ .

$3 x^{2} - 6 x + 0$

Step 1.1.1.3.2

Add $3 x^{2} - 6 x$ and $0$ .

$f' (x) = 3 x^{2} - 6 x$

Step 1.1.2

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} - 6 x$ .

$3 x^{2} - 6 x$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $3 x^{2} - 6 x = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$3 x^{2} - 6 x = 0$

Step 1.2.2

Factor $3 x$ out of $3 x^{2} - 6 x$ .

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Step 1.2.2.1

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

$3 x (x) - 6 x = 0$

Step 1.2.2.2

Factor $3 x$ out of $- 6 x$ .

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

Step 1.2.2.3

Factor $3 x$ out of $3 x (x) + 3 x (- 2)$ .

$3 x (x - 2) = 0$

Step 1.2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 2 = 0$

Step 1.2.4

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.5

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

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Step 1.2.5.1

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

$x - 2 = 0$

Step 1.2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.2.6

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

$x = 0, 2$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

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 1.4

Evaluate $x^{3} - 3 x^{2} + 12$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = 0$ .

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Step 1.4.1.1

Substitute $0$ for $x$ .

${(0)}^{3} - 3 {(0)}^{2} + 12$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

Raising $0$ to any positive power yields $0$ .

$0 - 3 {(0)}^{2} + 12$

Step 1.4.1.2.1.2

Raising $0$ to any positive power yields $0$ .

$0 - 3 \cdot 0 + 12$

Step 1.4.1.2.1.3

Multiply $- 3$ by $0$ .

$0 + 0 + 12$

Step 1.4.1.2.2

Simplify by adding numbers.

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Step 1.4.1.2.2.1

Add $0$ and $0$ .

$0 + 12$

Step 1.4.1.2.2.2

Add $0$ and $12$ .

$12$

Step 1.4.2

Evaluate at $x = 2$ .

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Step 1.4.2.1

Substitute $2$ for $x$ .

${(2)}^{3} - 3 {(2)}^{2} + 12$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

Simplify each term.

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Step 1.4.2.2.1.1

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

$8 - 3 {(2)}^{2} + 12$

Step 1.4.2.2.1.2

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

$8 - 3 \cdot 4 + 12$

Step 1.4.2.2.1.3

Multiply $- 3$ by $4$ .

$8 - 12 + 12$

Step 1.4.2.2.2

Simplify by adding and subtracting.

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Step 1.4.2.2.2.1

Subtract $12$ from $8$ .

$- 4 + 12$

Step 1.4.2.2.2.2

Add $- 4$ and $12$ .

$8$

Step 1.4.3

List all of the points.

$(0, 12), (2, 8)$

Step 2

Use the first derivative test to determine which points can be maxima or minima.

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Step 2.1

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

$(- \infty, 0) \cup (0, 2) \cup (2, \infty)$

Step 2.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $3 x^{2} - 6 x$ to check if the result is negative or positive.

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Step 2.2.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = 3 {(- 2)}^{2} - 6 \cdot - 2$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

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

$f' (- 2) = 3 \cdot 4 - 6 \cdot - 2$

Step 2.2.2.1.2

Multiply $3$ by $4$ .

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

Step 2.2.2.1.3

Multiply $- 6$ by $- 2$ .

$f' (- 2) = 12 + 12$

Step 2.2.2.2

Add $12$ and $12$ .

$f' (- 2) = 24$

Step 2.2.2.3

The final answer is $24$ .

$24$

Step 2.3

Substitute any number, such as $1$ , from the interval $(0, 2)$ in the first derivative $3 x^{2} - 6 x$ to check if the result is negative or positive.

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Step 2.3.1

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

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

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

One to any power is one.

$f' (1) = 3 \cdot 1 - 6 \cdot 1$

Step 2.3.2.1.2

Multiply $3$ by $1$ .

$f' (1) = 3 - 6 \cdot 1$

Step 2.3.2.1.3

Multiply $- 6$ by $1$ .

$f' (1) = 3 - 6$

Step 2.3.2.2

Subtract $6$ from $3$ .

$f' (1) = - 3$

Step 2.3.2.3

The final answer is $- 3$ .

$- 3$

Step 2.4

Substitute any number, such as $4$ , from the interval $(2, \infty)$ in the first derivative $3 x^{2} - 6 x$ to check if the result is negative or positive.

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Step 2.4.1

Replace the variable $x$ with $4$ in the expression.

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

Step 2.4.2

Simplify the result.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

Raise $4$ to the power of $2$ .

$f' (4) = 3 \cdot 16 - 6 \cdot 4$

Step 2.4.2.1.2

Multiply $3$ by $16$ .

$f' (4) = 48 - 6 \cdot 4$

Step 2.4.2.1.3

Multiply $- 6$ by $4$ .

$f' (4) = 48 - 24$

Step 2.4.2.2

Subtract $24$ from $48$ .

$f' (4) = 24$

Step 2.4.2.3

The final answer is $24$ .

$24$

Step 2.5

Since the first derivative changed signs from positive to negative around $x = 0$ , then $x = 0$ is a local maximum.

$x = 0$ is a local maximum

Step 2.6

Since the first derivative changed signs from negative to positive around $x = 2$ , then $x = 2$ is a local minimum.

$x = 2$ is a local minimum

Step 2.7

These are the local extrema for $f (x) = x^{3} - 3 x^{2} + 12$ .

$x = 0$ is a local maximum

$x = 2$ is a local minimum

$x = 0$ is a local maximum

$x = 2$ is a local minimum

Step 3

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

Absolute Maximum: $(0, 12)$

Absolute Minimum: $(2, 8)$

Step 4