Calculus Examples

h (x) = {\begin{matrix} - x^{2} + k^{2}, 0 \leq x \leq 1 \frac{4 x + 4}{2 - x}, x > 1 \end{matrix}

$h (x) = {\begin{cases} - x^{2} + k^{2}, 0 \leq x \leq 1 \\ \frac{4 x + 4}{2 - x}, x > 1 \end{cases}$

Step 1

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Step 1.1

To find whether the function is continuous on

(1, \infty)

$(1, \infty)$ or not, find the domain of

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

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

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Step 1.1.1

Set the denominator in

\frac{4 x + 4}{2 - x}

$\frac{4 x + 4}{2 - x}$ equal to

0

$0$ to find where the expression is undefined.

2 - x = 0

$2 - x = 0$

Step 1.1.2

Solve for

x

$x$ .

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Step 1.1.2.1

Subtract

2

$2$ from both sides of the equation.

- x = - 2

$- x = - 2$

Step 1.1.2.2

Divide each term in

- x = - 2

$- x = - 2$ by

- 1

$- 1$ and simplify.

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Step 1.1.2.2.1

Divide each term in

- x = - 2

$- x = - 2$ by

- 1

$- 1$ .

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

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

Step 1.1.2.2.2

Simplify the left side.

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Step 1.1.2.2.2.1

Dividing two negative values results in a positive value.

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

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

Step 1.1.2.2.2.2

Divide

x

$x$ by

1

$1$ .

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

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

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

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

Step 1.1.2.2.3

Simplify the right side.

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Step 1.1.2.2.3.1

Divide

- 2

$- 2$ by

- 1

$- 1$ .

x = 2

$x = 2$

x = 2

$x = 2$

x = 2

$x = 2$

x = 2

$x = 2$

Step 1.1.3

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

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

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

Set-Builder Notation:

{x | x \neq 2}

${x | x \neq 2}$

Interval Notation:

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

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

Set-Builder Notation:

{x | x \neq 2}

${x | x \neq 2}$

Step 1.2

f (x)

$f (x)$ is not continuous on

(1, \infty)

$(1, \infty)$ because

2

$2$ is not in the domain of

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

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

The function is not continuous.

Step 2

Not continuous

Step 3