Calculus Examples

$f (x) = \frac{1}{x}$ , $[2, 6]$

Find the derivative.

Tap for more steps...

Find the first derivative.

Tap for more steps...

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$f' (x) = \frac{d}{d x} (x^{- 1})$

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

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

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = - \frac{1}{x^{2}}$

The first derivative of $f (x)$ with respect to $x$ is $- \frac{1}{x^{2}}$ .

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

Find if the derivative is continuous on $[2, 6]$ .

Tap for more steps...

To find whether the function is continuous on $[2, 6]$ or not, find the domain of $f' (x) = - \frac{1}{x^{2}}$ .

Tap for more steps...

Set the denominator in $\frac{1}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Solve for $x$ .

Tap for more steps...

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Tap for more steps...

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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:

${x | x \neq 0}$

$f' (x)$ is continuous on $[2, 6]$ .

The function is continuous.

The function is differentiable on $[2, 6]$ because the derivative is continuous on $[2, 6]$ .

The function is differentiable.

Enter YOUR Problem