Calculus Examples

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

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

Differentiate.

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Step 1.1.1.1

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

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

Step 1.1.1.2

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

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

Step 1.1.2

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

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Step 1.1.2.1

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

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

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 $n = 1$ .

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

Step 1.1.2.3

Multiply $- 2$ by $1$ .

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

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

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

$2 x - 2 + 0$

Step 1.1.3.2

Add $2 x - 2$ and $0$ .

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

Step 1.2

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

$2 x - 2$

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

$2 x - 2 = 0$

Step 2.2

Add $2$ to both sides of the equation.

$2 x = 2$

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

Divide $x$ by $1$ .

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $2$ by $2$ .

$x = 1$

Step 3

The values which make the derivative equal to $0$ are $1$ .

$1$

Step 4

After finding the point that makes the derivative $f' (x) = 2 x - 2$ equal to $0$ or undefined, the interval to check where $f (x) = x^{2} - 2 x + 1$ is increasing and where it is decreasing is $(- \infty, 1) \cup (1, \infty)$ .

$(- \infty, 1) \cup (1, \infty)$

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 $0$ in the expression.

$f' (0) = 2 (0) - 2$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $2$ by $0$ .

$f' (0) = 0 - 2$

Step 5.2.2

Subtract $2$ from $0$ .

$f' (0) = - 2$

Step 5.2.3

The final answer is $- 2$ .

$- 2$

Step 5.3

At $x = 0$ the derivative is $- 2$ . Since this is negative, the function is decreasing on $(- \infty, 1)$ .

Decreasing on $(- \infty, 1)$ since $f' (x) < 0$

Step 6

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

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Step 6.1

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

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

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $2$ .

$f' (2) = 4 - 2$

Step 6.2.2

Subtract $2$ from $4$ .

$f' (2) = 2$

Step 6.2.3

The final answer is $2$ .

$2$

Step 6.3

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

Increasing on $(1, \infty)$ since $f' (x) > 0$

Step 7

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

Increasing on: $(1, \infty)$

Decreasing on: $(- \infty, 1)$

Step 8