Calculus Examples

f (x) = \frac{x^{2} - 64}{x}

$f (x) = \frac{x^{2} - 64}{x}$ ,

[- 8, 8]

$[- 8, 8]$

Step 1

f

$f$ is continuous on the interval

[a, b]

$[a, b]$ and differentiable on

(a, b)

$(a, b)$ , then at least one real number

c

$c$ exists in the interval

(a, b)

$(a, b)$ such that

f' (c) = \frac{f (b) - f a}{b - a}

$f' (c) = \frac{f (b) - f a}{b - a}$ . The mean value theorem expresses the relationship between the slope of the tangent to the curve at

x = c

$x = c$ and the slope of the line through the points

(a, f (a))

$(a, f (a))$ and

(b, f (b))

$(b, f (b))$ .

f (x)

$f (x)$ is continuous on

[a, b]

$[a, b]$

and if

f (x)

$f (x)$ differentiable on

(a, b)

$(a, b)$ ,

then there exists at least one point,

c

$c$ in

[a, b]

$[a, b]$ :

f' (c) = \frac{f (b) - f a}{b - a}

$f' (c) = \frac{f (b) - f a}{b - a}$ .

Step 2

Check if

f (x) = \frac{x^{2} - 64}{x}

$f (x) = \frac{x^{2} - 64}{x}$ is continuous.

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Step 2.1

To find whether the function is continuous on

[- 8, 8]

$[- 8, 8]$ or not, find the domain of

f (x) = \frac{x^{2} - 64}{x}

$f (x) = \frac{x^{2} - 64}{x}$ .

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Step 2.1.1

Set the denominator in

\frac{x^{2} - 64}{x}

$\frac{x^{2} - 64}{x}$ equal to

0

$0$ to find where the expression is undefined.

x = 0

$x = 0$

Step 2.1.2

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

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

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

Set-Builder Notation:

{x | x \neq 0}

${x | x \neq 0}$

Interval Notation:

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

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

Set-Builder Notation:

{x | x \neq 0}

${x | x \neq 0}$

Step 2.2

f (x)

$f (x)$ is not continuous on

[- 8, 8]

$[- 8, 8]$ because

0

$0$ is not in the domain of

f (x) = \frac{x^{2} - 64}{x}

$f (x) = \frac{x^{2} - 64}{x}$ .

The function is not continuous.

Step 3