Calculus Examples

$f (x) = x (6 - x)$ ; $(- \infty, \infty)$

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 using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = 6 - x$ .

$x \frac{d}{d x} [6 - x] + (6 - x) \frac{d}{d x} [x]$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

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

$x (\frac{d}{d x} [6] + \frac{d}{d x} [- x]) + (6 - x) \frac{d}{d x} [x]$

Step 1.1.1.2.2

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

$x (0 + \frac{d}{d x} [- x]) + (6 - x) \frac{d}{d x} [x]$

Step 1.1.1.2.3

Add $0$ and $\frac{d}{d x} [- x]$ .

$x \frac{d}{d x} [- x] + (6 - x) \frac{d}{d x} [x]$

Step 1.1.1.2.4

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

$x (- \frac{d}{d x} [x]) + (6 - x) \frac{d}{d x} [x]$

Step 1.1.1.2.5

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

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

Step 1.1.1.2.6

Simplify the expression.

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Step 1.1.1.2.6.1

Multiply $- 1$ by $1$ .

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

Step 1.1.1.2.6.2

Move $- 1$ to the left of $x$ .

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

Step 1.1.1.2.6.3

Rewrite $- 1 x$ as $- x$ .

$- x + (6 - x) \frac{d}{d x} [x]$

Step 1.1.1.2.7

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

$- x + (6 - x) \cdot 1$

Step 1.1.1.2.8

Simplify by adding terms.

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Step 1.1.1.2.8.1

Multiply $6 - x$ by $1$ .

$- x + 6 - x$

Step 1.1.1.2.8.2

Subtract $x$ from $- x$ .

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

Step 1.1.2

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

$- 2 x + 6$

Step 1.2

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

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Step 1.2.1

Set the first derivative equal to $0$ .

$- 2 x + 6 = 0$

Step 1.2.2

Subtract $6$ from both sides of the equation.

$- 2 x = - 6$

Step 1.2.3

Divide each term in $- 2 x = - 6$ by $- 2$ and simplify.

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Step 1.2.3.1

Divide each term in $- 2 x = - 6$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 6}{- 2}$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $- 2$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 6}{- 2}$

Step 1.2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 6}{- 2}$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Divide $- 6$ by $- 2$ .

$x = 3$

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 $x (6 - x)$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = 3$ .

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Step 1.4.1.1

Substitute $3$ for $x$ .

$(3) (6 - (3))$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Multiply $- 1$ by $3$ .

$3 (6 - 3)$

Step 1.4.1.2.2

Subtract $3$ from $6$ .

$3 \cdot 3$

Step 1.4.1.2.3

Multiply $3$ by $3$ .

$9$

Step 1.4.2

List all of the points.

$(3, 9)$

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.

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

Step 2.2

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

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Step 2.2.1

Replace the variable $x$ with $0$ in the expression.

$f' (0) = - 2 \cdot 0 + 6$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Multiply $- 2$ by $0$ .

$f' (0) = 0 + 6$

Step 2.2.2.2

Add $0$ and $6$ .

$f' (0) = 6$

Step 2.2.2.3

The final answer is $6$ .

$6$

Step 2.3

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

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Step 2.3.1

Replace the variable $x$ with $6$ in the expression.

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

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Multiply $- 2$ by $6$ .

$f' (6) = - 12 + 6$

Step 2.3.2.2

Add $- 12$ and $6$ .

$f' (6) = - 6$

Step 2.3.2.3

The final answer is $- 6$ .

$- 6$

Step 2.4

Since the first derivative changed signs from positive to negative around $x = 3$ , then $x = 3$ is a local maximum.

$x = 3$ is a local maximum

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: $(3, 9)$

No absolute minimum

Step 4