Calculus Examples

$f (x) = 5 x^{3} - 5 x^{2}$

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 $5 x^{3} - 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [5 x^{3}] + \frac{d}{d x} [- 5 x^{2}]$ .

$\frac{d}{d x} [5 x^{3}] + \frac{d}{d x} [- 5 x^{2}]$

Step 1.1.2

Evaluate $\frac{d}{d x} [5 x^{3}]$ .

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Step 1.1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{3}$ with respect to $x$ is $5 \frac{d}{d x} [x^{3}]$ .

$5 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 x^{2}]$

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 $n = 3$ .

$5 (3 x^{2}) + \frac{d}{d x} [- 5 x^{2}]$

Step 1.1.2.3

Multiply $3$ by $5$ .

$15 x^{2} + \frac{d}{d x} [- 5 x^{2}]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 5 x^{2}]$ .

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Step 1.1.3.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{2}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{2}]$ .

$15 x^{2} - 5 \frac{d}{d x} [x^{2}]$

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 $n = 2$ .

$15 x^{2} - 5 (2 x)$

Step 1.1.3.3

Multiply $2$ by $- 5$ .

$f' (x) = 15 x^{2} - 10 x$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $15 x^{2} - 10 x$ with respect to $x$ is $\frac{d}{d x} [15 x^{2}] + \frac{d}{d x} [- 10 x]$ .

$\frac{d}{d x} [15 x^{2}] + \frac{d}{d x} [- 10 x]$

Step 1.2.2

Evaluate $\frac{d}{d x} [15 x^{2}]$ .

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Step 1.2.2.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x^{2}$ with respect to $x$ is $15 \frac{d}{d x} [x^{2}]$ .

$15 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 10 x]$

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 $n = 2$ .

$15 (2 x) + \frac{d}{d x} [- 10 x]$

Step 1.2.2.3

Multiply $2$ by $15$ .

$30 x + \frac{d}{d x} [- 10 x]$

Step 1.2.3

Evaluate $\frac{d}{d x} [- 10 x]$ .

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Step 1.2.3.1

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x$ with respect to $x$ is $- 10 \frac{d}{d x} [x]$ .

$30 x - 10 \frac{d}{d x} [x]$

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 $n = 1$ .

$30 x - 10 \cdot 1$

Step 1.2.3.3

Multiply $- 10$ by $1$ .

$f'' (x) = 30 x - 10$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $30 x - 10$ .

$30 x - 10$

Step 2

Set the second derivative equal to $0$ then solve the equation $30 x - 10 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$30 x - 10 = 0$

Step 2.2

Add $10$ to both sides of the equation.

$30 x = 10$

Step 2.3

Divide each term in $30 x = 10$ by $30$ and simplify.

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Step 2.3.1

Divide each term in $30 x = 10$ by $30$ .

$\frac{30 x}{30} = \frac{10}{30}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $30$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{30 x}{30} = \frac{10}{30}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{10}{30}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $10$ and $30$ .

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Step 2.3.3.1.1

Factor $10$ out of $10$ .

$x = \frac{10 (1)}{30}$

Step 2.3.3.1.2

Cancel the common factors.

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Step 2.3.3.1.2.1

Factor $10$ out of $30$ .

$x = \frac{10 \cdot 1}{10 \cdot 3}$

Step 2.3.3.1.2.2

Cancel the common factor.

$x = \frac{10 \cdot 1}{10 \cdot 3}$

Step 2.3.3.1.2.3

Rewrite the expression.

$x = \frac{1}{3}$

Step 3

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

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Step 3.1

Substitute $\frac{1}{3}$ in $f (x) = 5 x^{3} - 5 x^{2}$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $\frac{1}{3}$ in the expression.

$f (\frac{1}{3}) = 5 {(\frac{1}{3})}^{3} - 5 {(\frac{1}{3})}^{2}$

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

Apply the product rule to $\frac{1}{3}$ .

$f (\frac{1}{3}) = 5 (\frac{1^{3}}{3^{3}}) - 5 {(\frac{1}{3})}^{2}$

Step 3.1.2.1.2

One to any power is one.

$f (\frac{1}{3}) = 5 (\frac{1}{3^{3}}) - 5 {(\frac{1}{3})}^{2}$

Step 3.1.2.1.3

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

$f (\frac{1}{3}) = 5 (\frac{1}{27}) - 5 {(\frac{1}{3})}^{2}$

Step 3.1.2.1.4

Combine $5$ and $\frac{1}{27}$ .

$f (\frac{1}{3}) = \frac{5}{27} - 5 {(\frac{1}{3})}^{2}$

Step 3.1.2.1.5

Apply the product rule to $\frac{1}{3}$ .

$f (\frac{1}{3}) = \frac{5}{27} - 5 \frac{1^{2}}{3^{2}}$

Step 3.1.2.1.6

One to any power is one.

$f (\frac{1}{3}) = \frac{5}{27} - 5 \frac{1}{3^{2}}$

Step 3.1.2.1.7

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

$f (\frac{1}{3}) = \frac{5}{27} - 5 (\frac{1}{9})$

Step 3.1.2.1.8

Combine $- 5$ and $\frac{1}{9}$ .

$f (\frac{1}{3}) = \frac{5}{27} + \frac{- 5}{9}$

Step 3.1.2.1.9

Move the negative in front of the fraction.

$f (\frac{1}{3}) = \frac{5}{27} - \frac{5}{9}$

Step 3.1.2.2

To write $- \frac{5}{9}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{5}{27} - \frac{5}{9} \cdot \frac{3}{3}$

Step 3.1.2.3

Write each expression with a common denominator of $27$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.1.2.3.1

Multiply $\frac{5}{9}$ by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{5}{27} - \frac{5 \cdot 3}{9 \cdot 3}$

Step 3.1.2.3.2

Multiply $9$ by $3$ .

$f (\frac{1}{3}) = \frac{5}{27} - \frac{5 \cdot 3}{27}$

Step 3.1.2.4

Combine the numerators over the common denominator.

$f (\frac{1}{3}) = \frac{5 - 5 \cdot 3}{27}$

Step 3.1.2.5

Simplify the numerator.

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Step 3.1.2.5.1

Multiply $- 5$ by $3$ .

$f (\frac{1}{3}) = \frac{5 - 15}{27}$

Step 3.1.2.5.2

Subtract $15$ from $5$ .

$f (\frac{1}{3}) = \frac{- 10}{27}$

Step 3.1.2.6

Move the negative in front of the fraction.

$f (\frac{1}{3}) = - \frac{10}{27}$

Step 3.1.2.7

The final answer is $- \frac{10}{27}$ .

$- \frac{10}{27}$

Step 3.2

The point found by substituting $\frac{1}{3}$ in $f (x) = 5 x^{3} - 5 x^{2}$ is $(\frac{1}{3}, - \frac{10}{27})$ . This point can be an inflection point.

$(\frac{1}{3}, - \frac{10}{27})$

Step 4

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

$(- \infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{1}{3})$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $0.2\overline{3}$ in the expression.

$f'' (0.2\overline{3}) = 30 (0.2\overline{3}) - 10$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $30$ by $0.2\overline{3}$ .

$f'' (0.2\overline{3}) = 7 - 10$

Step 5.2.2

Subtract $10$ from $7$ .

$f'' (0.2\overline{3}) = - 3$

Step 5.2.3

The final answer is $- 3$ .

$- 3$

Step 5.3

At $0.2\overline{3}$ , the second derivative is $- 3$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{1}{3})$

Decreasing on $(- \infty, \frac{1}{3})$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(\frac{1}{3}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.4\overline{3}$ in the expression.

$f'' (0.4\overline{3}) = 30 (0.4\overline{3}) - 10$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $30$ by $0.4\overline{3}$ .

$f'' (0.4\overline{3}) = 13 - 10$

Step 6.2.2

Subtract $10$ from $13$ .

$f'' (0.4\overline{3}) = 3$

Step 6.2.3

The final answer is $3$ .

$3$

Step 6.3

At $0.4\overline{3}$ , the second derivative is $3$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{3}, \infty)$ .

Increasing on $(\frac{1}{3}, \infty)$ 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 $(\frac{1}{3}, - \frac{10}{27})$ .

$(\frac{1}{3}, - \frac{10}{27})$

Step 8

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