Calculus Examples

$f (x) = x^{2} - 4 x$

Find the first derivative.

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Find the first derivative.

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Differentiate.

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

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

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

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

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

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

$f' (x) = 2 x - 4 \frac{d}{d x} x$

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

$f' (x) = 2 x - 4 \cdot 1$

Multiply $- 4$ by $1$ .

$f' (x) = 2 x - 4$

The first derivative of $f (x)$ with respect to $x$ is $2 x - 4$ .

$2 x - 4$

Set the first derivative equal to $0$ then solve the equation $2 x - 4 = 0$ .

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Set the first derivative equal to $0$ .

$2 x - 4 = 0$

Add $4$ to both sides of the equation.

$2 x = 4$

Divide each term in $2 x = 4$ by $2$ and simplify.

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Divide each term in $2 x = 4$ by $2$ .

$\frac{2 x}{2} = \frac{4}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{4}{2}$

Divide $x$ by $1$ .

$x = \frac{4}{2}$

Simplify the right side.

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Divide $4$ by $2$ .

$x = 2$

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

$2$

After finding the point that makes the derivative $f' (x) = 2 x - 4$ equal to $0$ or undefined, the interval to check where $f (x) = x^{2} - 4 x$ is increasing and where it is decreasing is $(- \infty, 2) \cup (2, \infty)$ .

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

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

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

$f' (1) = 2 (1) - 4$

Simplify the result.

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Multiply $2$ by $1$ .

$f' (1) = 2 - 4$

Subtract $4$ from $2$ .

$f' (1) = - 2$

The final answer is $- 2$ .

$- 2$

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

Decreasing on $(- \infty, 2)$ since $f' (x) < 0$

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

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Replace the variable $x$ with $3$ in the expression.

$f' (3) = 2 (3) - 4$

Simplify the result.

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Multiply $2$ by $3$ .

$f' (3) = 6 - 4$

Subtract $4$ from $6$ .

$f' (3) = 2$

The final answer is $2$ .

$2$

At $x = 3$ the derivative is $2$ . Since this is positive, the function is increasing on $(2, \infty)$ .

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

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

Increasing on: $(2, \infty)$

Decreasing on: $(- \infty, 2)$