Calculus Examples

$f (x) = \frac{x^{2} - 64}{x}$ , $[- 8, 8]$

Step 1

If $f$ is continuous on the interval $[a, b]$ and differentiable on $(a, b)$ , then at least one real number $c$ exists in the interval $(a, b)$ such that $f' (c) = \frac{f (b) - f a}{b - a}$ . The mean value theorem expresses the relationship between the slope of the tangent to the curve at $x = c$ and the slope of the line through the points $(a, f (a))$ and $(b, f (b))$ .

If $f (x)$ is continuous on $[a, b]$

and if $f (x)$ differentiable on $(a, b)$ ,

then there exists at least one point, $c$ in $[a, b]$ : $f' (c) = \frac{f (b) - f a}{b - a}$ .

Step 2

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

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Step 2.1

To find whether the function is continuous on $[- 8, 8]$ or not, find the domain of $f (x) = \frac{x^{2} - 64}{x}$ .

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Step 2.1.1

Set the denominator in $\frac{x^{2} - 64}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 2.1.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$

Step 2.2

$f (x)$ is not continuous on $[- 8, 8]$ because $0$ is not in the domain of $f (x) = \frac{x^{2} - 64}{x}$ .

The function is not continuous.

Step 3