Calculus Examples

$f (x) = - x^{5}$

Find the inflection points.

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Find the second derivative.

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Find the first derivative.

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{5}$ with respect to $x$ is $- \frac{d}{d x} [x^{5}]$ .

$f' (x) = - \frac{d}{d x} x^{5}$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f' (x) = - (5 x^{4})$

Multiply $5$ by $- 1$ .

$f' (x) = - 5 x^{4}$

Find the second derivative.

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Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{4}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = - 5 \frac{d}{d x} x^{4}$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = - 5 (4 x^{3})$

Multiply $4$ by $- 5$ .

$f'' (x) = - 20 x^{3}$

The second derivative of $f (x)$ with respect to $x$ is $- 20 x^{3}$ .

$- 20 x^{3}$

Set the second derivative equal to $0$ then solve the equation $- 20 x^{3} = 0$ .

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Set the second derivative equal to $0$ .

$- 20 x^{3} = 0$

Divide each term by $- 20$ and simplify.

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Divide each term in $- 20 x^{3} = 0$ by $- 20$ .

$\frac{- 20 x^{3}}{- 20} = \frac{0}{- 20}$

Cancel the common factor of $- 20$ .

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Cancel the common factor.

$\frac{- 20 x^{3}}{- 20} = \frac{0}{- 20}$

Divide $x^{3}$ by $1$ .

$x^{3} = \frac{0}{- 20}$

Divide $0$ by $- 20$ .

$x^{3} = 0$

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

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Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

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

$x = 0$

Find the points where the second derivative is $0$ .

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Substitute $0$ in $f (x) = - x^{5}$ to find the value of $y$ .

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Replace the variable $x$ with $0$ in the expression.

$f (0) = - {(0)}^{5}$

Simplify the result.

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Raising $0$ to any positive power yields $0$ .

$f (0) = - 0$

Multiply $- 1$ by $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

The point found by substituting $0$ in $f (x) = - x^{5}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

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

Substitute a value from the interval $(- \infty, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = - 20 {(- 0.1)}^{3}$

Simplify the result.

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Raise $- 0.1$ to the power of $3$ .

$f'' (- 0.1) = - 20 \cdot - 0.001$

Multiply $- 20$ by $- 0.001$ .

$f'' (- 0.1) = 0.02$

The final answer is $0.02$ .

$0.02$

At $- 0.1$ , the second derivative is $0.02$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f'' (x) > 0$

Substitute a value from the interval $(0, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = - 20 {(0.1)}^{3}$

Simplify the result.

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Raise $0.1$ to the power of $3$ .

$f'' (0.1) = - 20 \cdot 0.001$

Multiply $- 20$ by $0.001$ .

$f'' (0.1) = - 0.02$

The final answer is $- 0.02$ .

$- 0.02$

At $0.1$ , the second derivative is $- 0.02$ . Since this is negative, the second derivative is decreasing on the interval $(0, \infty)$

Decreasing on $(0, \infty)$ since $f'' (x) < 0$

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(0, 0)$ .

$(0, 0)$

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Create intervals around the inflection points and the undefined values.

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

Substitute any number from the interval $(- \infty, 0)$ into the second derivative and evaluate to determine the concavity.

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Replace the variable $x$ with $- 2$ in the expression.

$f'' (- 2) = - 20 {(- 2)}^{3}$

Simplify the result.

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Raise $- 2$ to the power of $3$ .

$f'' (- 2) = - 20 \cdot - 8$

Multiply $- 20$ by $- 8$ .

$f'' (- 2) = 160$

The final answer is $160$ .

$160$

The graph is concave up on the interval $(- \infty, 0)$ because $f'' (- 2)$ is positive.

Concave up on $(- \infty, 0)$ since $f'' (x)$ is positive

Substitute any number from the interval $(0, \infty)$ into the second derivative and evaluate to determine the concavity.

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Replace the variable $x$ with $2$ in the expression.

$f'' (2) = - 20 {(2)}^{3}$

Simplify the result.

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Raise $2$ to the power of $3$ .

$f'' (2) = - 20 \cdot 8$

Multiply $- 20$ by $8$ .

$f'' (2) = - 160$

The final answer is $- 160$ .

$- 160$

The graph is concave down on the interval $(0, \infty)$ because $f'' (2)$ is negative.

Concave down on $(0, \infty)$ since $f'' (x)$ is negative

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on $(- \infty, 0)$ since $f'' (x)$ is positive

Concave down on $(0, \infty)$ since $f'' (x)$ is negative

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