Calculus Examples

$f (x) = x^{3}$

Step 1

Find the second derivative.

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Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 3 x^{2}$

Find the second derivative.

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Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 3 \frac{d}{d x} (x^{2})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 3 (2 x)$

Multiply $2$ by $3$ .

$f'' (x) = 6 x$

The second derivative of $f (x)$ with respect to $x$ is $6 x$ .

$6 x$

Step 2

Set the second derivative equal to $0$ then solve the equation $6 x = 0$ .

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Set the second derivative equal to $0$ .

$6 x = 0$

Divide each term in $6 x = 0$ by $6$ and simplify.

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Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Simplify the left side.

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Cancel the common factor of $6$ .

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Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Simplify the right side.

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Divide $0$ by $6$ .

$x = 0$

Step 3

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

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Substitute $0$ in $f (x) = x^{3}$ to find the value of $y$ .

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

$f (0) = {(0)}^{3}$

Simplify the result.

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

$f (0) = 0$

The final answer is $0$ .

$0$

The point found by substituting $0$ in $f (x) = x^{3}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4

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

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

Step 5

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.

$f'' (- 0.1) = 6 (- 0.1)$

Simplify the result.

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Multiply $6$ by $- 0.1$ .

$f'' (- 0.1) = - 0.6$

The final answer is $- 0.6$ .

$- 0.6$

At $- 0.1$ , the second derivative is $- 0.6$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 0)$

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

Step 6

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.

$f'' (0.1) = 6 (0.1)$

Simplify the result.

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Multiply $6$ by $0.1$ .

$f'' (0.1) = 0.6$

The final answer is $0.6$ .

$0.6$

At $0.1$ , the second derivative is $0.6$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

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

Step 7

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 $(0, 0)$ .

$(0, 0)$

Step 8