Calculus Examples

$g (x) = x (x^{4} - 3)$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = x^{4} - 3$ .

$x \frac{d}{d x} [x^{4} - 3] + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $x^{4} - 3$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3]$ .

$x (\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3]) + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.2.2

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

$x (4 x^{3} + \frac{d}{d x} [- 3]) + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.2.3

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$x (4 x^{3} + 0) + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.2.4

Add $4 x^{3}$ and $0$ .

$x (4 x^{3}) + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.3

Raise $x$ to the power of $1$ .

$4 (x^{1} x^{3}) + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 x^{1 + 3} + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.5

Add $1$ and $3$ .

$4 x^{4} + (x^{4} - 3) \frac{d}{d x} [x]$

Step 1.1.6

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

$4 x^{4} + (x^{4} - 3) \cdot 1$

Step 1.1.7

Simplify by adding terms.

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Step 1.1.7.1

Multiply $x^{4} - 3$ by $1$ .

$4 x^{4} + x^{4} - 3$

Step 1.1.7.2

Add $4 x^{4}$ and $x^{4}$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $5 x^{4} - 3$ with respect to $x$ is $\frac{d}{d x} [5 x^{4}] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [5 x^{4}] + \frac{d}{d x} [- 3]$

Step 1.2.2

Evaluate $\frac{d}{d x} [5 x^{4}]$ .

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Step 1.2.2.1

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

$5 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3]$

Step 1.2.2.2

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

$5 (4 x^{3}) + \frac{d}{d x} [- 3]$

Step 1.2.2.3

Multiply $4$ by $5$ .

$20 x^{3} + \frac{d}{d x} [- 3]$

Step 1.2.3

Differentiate using the Constant Rule.

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Step 1.2.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$20 x^{3} + 0$

Step 1.2.3.2

Add $20 x^{3}$ and $0$ .

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

Step 1.3

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

$20 x^{3}$

Step 2

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

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Step 2.1

Set the second derivative equal to $0$ .

$20 x^{3} = 0$

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $20$ .

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Step 2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.1.2

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

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $0$ by $20$ .

$x^{3} = 0$

Step 2.3

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

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

$x = 0$

Step 3

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

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Step 3.1

Substitute $0$ in $g (x) = x (x^{4} - 3)$ to find the value of $y$ .

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Step 3.1.1

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

$g (0) = (0) ({(0)}^{4} - 3)$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

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

$g (0) = 0 (0 - 3)$

Step 3.1.2.2

Subtract $3$ from $0$ .

$g (0) = 0 \cdot - 3$

Step 3.1.2.3

Multiply $0$ by $- 3$ .

$g (0) = 0$

Step 3.1.2.4

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $0$ in $g (x) = x (x^{4} - 3)$ 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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Step 5.1

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

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

Step 5.2

Simplify the result.

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Step 5.2.1

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

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

Step 5.2.2

Multiply $20$ by $- 0.001$ .

$g'' (- 0.1) = - 0.02$

Step 5.2.3

The final answer is $- 0.02$ .

$- 0.02$

Step 5.3

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

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

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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Step 6.1

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

$g'' (0.1) = 20 {(0.1)}^{3}$

Step 6.2

Simplify the result.

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Step 6.2.1

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

$g'' (0.1) = 20 \cdot 0.001$

Step 6.2.2

Multiply $20$ by $0.001$ .

$g'' (0.1) = 0.02$

Step 6.2.3

The final answer is $0.02$ .

$0.02$

Step 6.3

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

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

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