Calculus Examples

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

Find the derivative.

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Find the first derivative.

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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 $n = - 1$ .

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

Remove the negative exponent by rewriting $x^{- 2}$ as $\frac{1}{x^{2}}$ .

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

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 $[- 6, 6]$ .

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

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Solve to find the value of $x$ that makes the expression undefined.

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Set up the equation to solve for $x$ .

$x^{2} = 0$

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

$x = \pm \sqrt{0}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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

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

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

$x = \pm 0$

$\pm 0$ is equal to $0$ .

$x = 0$

For asymptotes and point discontinuity (denominator equals $0$ ), it is easier to find where the expression is undefined. These values are not part of the domain.

$x^{2} = 0$

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

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

${x | x \neq 0}$

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

${x | x \neq 0}$

$f' (x)$ is not continuous on $[- 6, 6]$ because $0$ is not in the domain of $f' (x) = - \frac{1}{x^{2}}$ .

The function is not continuous.

The function is not differentiable on $[- 6, 6]$ because the derivative $- \frac{1}{x^{2}}$ is not continuous on $[- 6, 6]$ .

The function is not differentiable.

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