Calculus Examples

$a (x) = 0.005 x^{3} - 0.06 x^{2} + 0.2 + 0.02$

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

By the Sum Rule, the derivative of $0.005 x^{3} - 0.06 x^{2} + 0.2 + 0.02$ with respect to $x$ is $\frac{d}{d x} [0.005 x^{3}] + \frac{d}{d x} [- 0.06 x^{2}] + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$ .

$\frac{d}{d x} [0.005 x^{3}] + \frac{d}{d x} [- 0.06 x^{2}] + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$

Step 1.1.2

Evaluate $\frac{d}{d x} [0.005 x^{3}]$ .

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Step 1.1.2.1

Since $0.005$ is constant with respect to $x$ , the derivative of $0.005 x^{3}$ with respect to $x$ is $0.005 \frac{d}{d x} [x^{3}]$ .

$0.005 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 0.06 x^{2}] + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$

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 $n = 3$ .

$0.005 (3 x^{2}) + \frac{d}{d x} [- 0.06 x^{2}] + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$

Step 1.1.2.3

Multiply $3$ by $0.005$ .

$0.015 x^{2} + \frac{d}{d x} [- 0.06 x^{2}] + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 0.06 x^{2}]$ .

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Step 1.1.3.1

Since $- 0.06$ is constant with respect to $x$ , the derivative of $- 0.06 x^{2}$ with respect to $x$ is $- 0.06 \frac{d}{d x} [x^{2}]$ .

$0.015 x^{2} - 0.06 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$

Step 1.1.3.2

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

$0.015 x^{2} - 0.06 (2 x) + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$

Step 1.1.3.3

Multiply $2$ by $- 0.06$ .

$0.015 x^{2} - 0.12 x + \frac{d}{d x} [0.2] + \frac{d}{d x} [0.02]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

Since $0.2$ is constant with respect to $x$ , the derivative of $0.2$ with respect to $x$ is $0$ .

$0.015 x^{2} - 0.12 x + 0 + \frac{d}{d x} [0.02]$

Step 1.1.4.2

Since $0.02$ is constant with respect to $x$ , the derivative of $0.02$ with respect to $x$ is $0$ .

$0.015 x^{2} - 0.12 x + 0 + 0$

Step 1.1.5

Combine terms.

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Step 1.1.5.1

Add $0.015 x^{2} - 0.12 x$ and $0$ .

$0.015 x^{2} - 0.12 x + 0$

Step 1.1.5.2

Add $0.015 x^{2} - 0.12 x$ and $0$ .

$f' (x) = 0.015 x^{2} - 0.12 x$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $0.015 x^{2} - 0.12 x$ with respect to $x$ is $\frac{d}{d x} [0.015 x^{2}] + \frac{d}{d x} [- 0.12 x]$ .

$\frac{d}{d x} [0.015 x^{2}] + \frac{d}{d x} [- 0.12 x]$

Step 1.2.2

Evaluate $\frac{d}{d x} [0.015 x^{2}]$ .

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Step 1.2.2.1

Since $0.015$ is constant with respect to $x$ , the derivative of $0.015 x^{2}$ with respect to $x$ is $0.015 \frac{d}{d x} [x^{2}]$ .

$0.015 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 0.12 x]$

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 $n = 2$ .

$0.015 (2 x) + \frac{d}{d x} [- 0.12 x]$

Step 1.2.2.3

Multiply $2$ by $0.015$ .

$0.03 x + \frac{d}{d x} [- 0.12 x]$

Step 1.2.3

Evaluate $\frac{d}{d x} [- 0.12 x]$ .

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Step 1.2.3.1

Since $- 0.12$ is constant with respect to $x$ , the derivative of $- 0.12 x$ with respect to $x$ is $- 0.12 \frac{d}{d x} [x]$ .

$0.03 x - 0.12 \frac{d}{d x} [x]$

Step 1.2.3.2

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

$0.03 x - 0.12 \cdot 1$

Step 1.2.3.3

Multiply $- 0.12$ by $1$ .

$f'' (x) = 0.03 x - 0.12$

Step 1.3

The second derivative of $a (x)$ with respect to $x$ is $0.03 x - 0.12$ .

$0.03 x - 0.12$

Step 2

Set the second derivative equal to $0$ then solve the equation $0.03 x - 0.12 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$0.03 x - 0.12 = 0$

Step 2.2

Add $0.12$ to both sides of the equation.

$0.03 x = 0.12$

Step 2.3

Divide each term in $0.03 x = 0.12$ by $0.03$ and simplify.

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Step 2.3.1

Divide each term in $0.03 x = 0.12$ by $0.03$ .

$\frac{0.03 x}{0.03} = \frac{0.12}{0.03}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $0.03$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{0.03 x}{0.03} = \frac{0.12}{0.03}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0.12}{0.03}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0.12$ by $0.03$ .

$x = 4$

Step 3

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

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Step 3.1

Substitute $4$ in $a (x) = 0.005 x^{3} - 0.06 x^{2} + 0.2 + 0.02$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $4$ in the expression.

$a (4) = 0.005 {(4)}^{3} - 0.06 {(4)}^{2} + 0.2 + 0.02$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Raise $4$ to the power of $3$ .

$a (4) = 0.005 \cdot 64 - 0.06 {(4)}^{2} + 0.2 + 0.02$

Step 3.1.2.1.2

Multiply $0.005$ by $64$ .

$a (4) = 0.32 - 0.06 {(4)}^{2} + 0.2 + 0.02$

Step 3.1.2.1.3

Raise $4$ to the power of $2$ .

$a (4) = 0.32 - 0.06 \cdot 16 + 0.2 + 0.02$

Step 3.1.2.1.4

Multiply $- 0.06$ by $16$ .

$a (4) = 0.32 - 0.96 + 0.2 + 0.02$

Step 3.1.2.2

Simplify by adding and subtracting.

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Step 3.1.2.2.1

Subtract $0.96$ from $0.32$ .

$a (4) = - 0.64 + 0.2 + 0.02$

Step 3.1.2.2.2

Add $- 0.64$ and $0.2$ .

$a (4) = - 0.44 + 0.02$

Step 3.1.2.2.3

Add $- 0.44$ and $0.02$ .

$a (4) = - 0.42$

Step 3.1.2.3

The final answer is $- 0.42$ .

$- 0.42$

Step 3.2

The point found by substituting $4$ in $a (x) = 0.005 x^{3} - 0.06 x^{2} + 0.2 + 0.02$ is $(4, - 0.42)$ . This point can be an inflection point.

$(4, - 0.42)$

Step 4

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

$(- \infty, 4) \cup (4, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 4)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $3.9$ in the expression.

$a'' (3.9) = 0.03 (3.9) - 0.12$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $0.03$ by $3.9$ .

$a'' (3.9) = 0.117 - 0.12$

Step 5.2.2

Subtract $0.12$ from $0.117$ .

$a'' (3.9) = - 0.003$

Step 5.2.3

The final answer is $- 0.003$ .

$- 0.003$

Step 5.3

At $3.9$ , the second derivative is $- 0.003$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 4)$

Decreasing on $(- \infty, 4)$ since $a'' (x) < 0$

Step 6

Substitute a value from the interval $(4, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $4.1$ in the expression.

$a'' (4.1) = 0.03 (4.1) - 0.12$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $0.03$ by $4.1$ .

$a'' (4.1) = 0.123 - 0.12$

Step 6.2.2

Subtract $0.12$ from $0.123$ .

$a'' (4.1) = 0.003$

Step 6.2.3

The final answer is $0.003$ .

$0.003$

Step 6.3

At $4.1$ , the second derivative is $0.003$ . Since this is positive, the second derivative is increasing on the interval $(4, \infty)$ .

Increasing on $(4, \infty)$ since $a'' (x) > 0$

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 $(4, - 0.42)$ .

$(4, - 0.42)$

Step 8