Calculus Examples

$4 x^{3} - 9 x^{4}$

Step 1

Write $4 x^{3} - 9 x^{4}$ as a function.

$f (x) = 4 x^{3} - 9 x^{4}$

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

By the Sum Rule, the derivative of $4 x^{3} - 9 x^{4}$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 9 x^{4}]$ .

$\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 9 x^{4}]$

Step 2.1.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 2.1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 9 x^{4}]$

Step 2.1.2.2

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

$4 (3 x^{2}) + \frac{d}{d x} [- 9 x^{4}]$

Step 2.1.2.3

Multiply $3$ by $4$ .

$12 x^{2} + \frac{d}{d x} [- 9 x^{4}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- 9 x^{4}]$ .

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Step 2.1.3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x^{4}$ with respect to $x$ is $- 9 \frac{d}{d x} [x^{4}]$ .

$12 x^{2} - 9 \frac{d}{d x} [x^{4}]$

Step 2.1.3.2

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

$12 x^{2} - 9 (4 x^{3})$

Step 2.1.3.3

Multiply $4$ by $- 9$ .

$12 x^{2} - 36 x^{3}$

Step 2.1.4

Reorder terms.

$f' (x) = - 36 x^{3} + 12 x^{2}$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $- 36 x^{3} + 12 x^{2}$ with respect to $x$ is $\frac{d}{d x} [- 36 x^{3}] + \frac{d}{d x} [12 x^{2}]$ .

$\frac{d}{d x} [- 36 x^{3}] + \frac{d}{d x} [12 x^{2}]$

Step 2.2.2

Evaluate $\frac{d}{d x} [- 36 x^{3}]$ .

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Step 2.2.2.1

Since $- 36$ is constant with respect to $x$ , the derivative of $- 36 x^{3}$ with respect to $x$ is $- 36 \frac{d}{d x} [x^{3}]$ .

$- 36 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [12 x^{2}]$

Step 2.2.2.2

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

$- 36 (3 x^{2}) + \frac{d}{d x} [12 x^{2}]$

Step 2.2.2.3

Multiply $3$ by $- 36$ .

$- 108 x^{2} + \frac{d}{d x} [12 x^{2}]$

Step 2.2.3

Evaluate $\frac{d}{d x} [12 x^{2}]$ .

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Step 2.2.3.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{2}$ with respect to $x$ is $12 \frac{d}{d x} [x^{2}]$ .

$- 108 x^{2} + 12 \frac{d}{d x} [x^{2}]$

Step 2.2.3.2

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

$- 108 x^{2} + 12 (2 x)$

Step 2.2.3.3

Multiply $2$ by $12$ .

$f'' (x) = - 108 x^{2} + 24 x$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $- 108 x^{2} + 24 x$ .

$- 108 x^{2} + 24 x$

Step 3

Set the second derivative equal to $0$ then solve the equation $- 108 x^{2} + 24 x = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$- 108 x^{2} + 24 x = 0$

Step 3.2

Factor $- 12 x$ out of $- 108 x^{2} + 24 x$ .

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Step 3.2.1

Factor $- 12 x$ out of $- 108 x^{2}$ .

$- 12 x (9 x) + 24 x = 0$

Step 3.2.2

Factor $- 12 x$ out of $24 x$ .

$- 12 x (9 x) - 12 x \cdot - 2 = 0$

Step 3.2.3

Factor $- 12 x$ out of $- 12 x (9 x) - 12 x (- 2)$ .

$- 12 x (9 x - 2) = 0$

Step 3.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$9 x - 2 = 0$

Step 3.4

Set $x$ equal to $0$ .

$x = 0$

Step 3.5

Set $9 x - 2$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $9 x - 2$ equal to $0$ .

$9 x - 2 = 0$

Step 3.5.2

Solve $9 x - 2 = 0$ for $x$ .

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Step 3.5.2.1

Add $2$ to both sides of the equation.

$9 x = 2$

Step 3.5.2.2

Divide each term in $9 x = 2$ by $9$ and simplify.

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Step 3.5.2.2.1

Divide each term in $9 x = 2$ by $9$ .

$\frac{9 x}{9} = \frac{2}{9}$

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $9$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{9 x}{9} = \frac{2}{9}$

Step 3.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{9}$

Step 3.6

The final solution is all the values that make $- 12 x (9 x - 2) = 0$ true.

$x = 0, \frac{2}{9}$

Step 4

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

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Step 4.1

Substitute $0$ in $f (x) = 4 x^{3} - 9 x^{4}$ to find the value of $y$ .

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Step 4.1.1

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

$f (0) = 4 {(0)}^{3} - 9 {(0)}^{4}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

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

$f (0) = 4 \cdot 0 - 9 {(0)}^{4}$

Step 4.1.2.1.2

Multiply $4$ by $0$ .

$f (0) = 0 - 9 {(0)}^{4}$

Step 4.1.2.1.3

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

$f (0) = 0 - 9 \cdot 0$

Step 4.1.2.1.4

Multiply $- 9$ by $0$ .

$f (0) = 0 + 0$

Step 4.1.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 4.1.2.3

The final answer is $0$ .

$0$

Step 4.2

The point found by substituting $0$ in $f (x) = 4 x^{3} - 9 x^{4}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4.3

Substitute $\frac{2}{9}$ in $f (x) = 4 x^{3} - 9 x^{4}$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $\frac{2}{9}$ in the expression.

$f (\frac{2}{9}) = 4 {(\frac{2}{9})}^{3} - 9 {(\frac{2}{9})}^{4}$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Apply the product rule to $\frac{2}{9}$ .

$f (\frac{2}{9}) = 4 (\frac{2^{3}}{9^{3}}) - 9 {(\frac{2}{9})}^{4}$

Step 4.3.2.1.2

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

$f (\frac{2}{9}) = 4 (\frac{8}{9^{3}}) - 9 {(\frac{2}{9})}^{4}$

Step 4.3.2.1.3

Raise $9$ to the power of $3$ .

$f (\frac{2}{9}) = 4 (\frac{8}{729}) - 9 {(\frac{2}{9})}^{4}$

Step 4.3.2.1.4

Multiply $4 (\frac{8}{729})$ .

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Step 4.3.2.1.4.1

Combine $4$ and $\frac{8}{729}$ .

$f (\frac{2}{9}) = \frac{4 \cdot 8}{729} - 9 {(\frac{2}{9})}^{4}$

Step 4.3.2.1.4.2

Multiply $4$ by $8$ .

$f (\frac{2}{9}) = \frac{32}{729} - 9 {(\frac{2}{9})}^{4}$

Step 4.3.2.1.5

Apply the product rule to $\frac{2}{9}$ .

$f (\frac{2}{9}) = \frac{32}{729} - 9 \frac{2^{4}}{9^{4}}$

Step 4.3.2.1.6

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

$f (\frac{2}{9}) = \frac{32}{729} - 9 \frac{16}{9^{4}}$

Step 4.3.2.1.7

Raise $9$ to the power of $4$ .

$f (\frac{2}{9}) = \frac{32}{729} - 9 (\frac{16}{6561})$

Step 4.3.2.1.8

Cancel the common factor of $9$ .

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Step 4.3.2.1.8.1

Factor $9$ out of $- 9$ .

$f (\frac{2}{9}) = \frac{32}{729} + 9 (- 1) (\frac{16}{6561})$

Step 4.3.2.1.8.2

Factor $9$ out of $6561$ .

$f (\frac{2}{9}) = \frac{32}{729} + 9 \cdot (- 1 \frac{16}{9 \cdot 729})$

Step 4.3.2.1.8.3

Cancel the common factor.

$f (\frac{2}{9}) = \frac{32}{729} + 9 \cdot (- 1 \frac{16}{9 \cdot 729})$

Step 4.3.2.1.8.4

Rewrite the expression.

$f (\frac{2}{9}) = \frac{32}{729} - 1 (\frac{16}{729})$

Step 4.3.2.1.9

Rewrite $- 1 (\frac{16}{729})$ as $- (\frac{16}{729})$ .

$f (\frac{2}{9}) = \frac{32}{729} - \frac{16}{729}$

Step 4.3.2.2

Combine fractions.

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Step 4.3.2.2.1

Combine the numerators over the common denominator.

$f (\frac{2}{9}) = \frac{32 - 16}{729}$

Step 4.3.2.2.2

Subtract $16$ from $32$ .

$f (\frac{2}{9}) = \frac{16}{729}$

Step 4.3.2.3

The final answer is $\frac{16}{729}$ .

$\frac{16}{729}$

Step 4.4

The point found by substituting $\frac{2}{9}$ in $f (x) = 4 x^{3} - 9 x^{4}$ is $(\frac{2}{9}, \frac{16}{729})$ . This point can be an inflection point.

$(\frac{2}{9}, \frac{16}{729})$

Step 4.5

Determine the points that could be inflection points.

$(0, 0), (\frac{2}{9}, \frac{16}{729})$

Step 5

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

$(- \infty, 0) \cup (0, \frac{2}{9}) \cup (\frac{2}{9}, \infty)$

Step 6

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 6.1

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

$f'' (- 0.1) = - 108 {(- 0.1)}^{2} + 24 (- 0.1)$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Raise $- 0.1$ to the power of $2$ .

$f'' (- 0.1) = - 108 \cdot 0.01 + 24 (- 0.1)$

Step 6.2.1.2

Multiply $- 108$ by $0.01$ .

$f'' (- 0.1) = - 1.08 + 24 (- 0.1)$

Step 6.2.1.3

Multiply $24$ by $- 0.1$ .

$f'' (- 0.1) = - 1.08 - 2.4$

Step 6.2.2

Subtract $2.4$ from $- 1.08$ .

$f'' (- 0.1) = - 3.48$

Step 6.2.3

The final answer is $- 3.48$ .

$- 3.48$

Step 6.3

At $- 0.1$ , the second derivative is $- 3.48$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 0)$

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

Step 7

Substitute a value from the interval $(0, \frac{2}{9})$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.\overline{1}$ in the expression.

$f'' (0.\overline{1}) = - 108 {(0.\overline{1})}^{2} + 24 (0.\overline{1})$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $0.\overline{1}$ to the power of $2$ .

$f'' (0.\overline{1}) = - 108 \cdot 0.01234567 + 24 (0.\overline{1})$

Step 7.2.1.2

Multiply $- 108$ by $0.01234567$ .

$f'' (0.\overline{1}) = - 1.33333333 + 24 (0.\overline{1})$

Step 7.2.1.3

Multiply $24$ by $0.\overline{1}$ .

$f'' (0.\overline{1}) = - 1.33333333 + 2.\overline{6}$

Step 7.2.2

Add $- 1.33333333$ and $2.\overline{6}$ .

$f'' (0.\overline{1}) = 1.33333333$

Step 7.2.3

The final answer is $1.33333333$ .

$1.33333333$

Step 7.3

At $0.\overline{1}$ , the second derivative is $1.33333333$ . Since this is positive, the second derivative is increasing on the interval $(0, \frac{2}{9})$ .

Increasing on $(0, \frac{2}{9})$ since $f'' (x) > 0$

Step 8

Substitute a value from the interval $(\frac{2}{9}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $0.3\overline{2}$ in the expression.

$f'' (0.3\overline{2}) = - 108 {(0.3\overline{2})}^{2} + 24 (0.3\overline{2})$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $0.3\overline{2}$ to the power of $2$ .

$f'' (0.3\overline{2}) = - 108 \cdot 0.10382716 + 24 (0.3\overline{2})$

Step 8.2.1.2

Multiply $- 108$ by $0.10382716$ .

$f'' (0.3\overline{2}) = - 11.21333333 + 24 (0.3\overline{2})$

Step 8.2.1.3

Multiply $24$ by $0.3\overline{2}$ .

$f'' (0.3\overline{2}) = - 11.21333333 + 7.73333333$

Step 8.2.2

Add $- 11.21333333$ and $7.73333333$ .

$f'' (0.3\overline{2}) = - 3.48$

Step 8.2.3

The final answer is $- 3.48$ .

$- 3.48$

Step 8.3

At $0.3\overline{2}$ , the second derivative is $- 3.48$ . Since this is negative, the second derivative is decreasing on the interval $(\frac{2}{9}, \infty)$

Decreasing on $(\frac{2}{9}, \infty)$ since $f'' (x) < 0$

Step 9

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 points in this case are $(0, 0), (\frac{2}{9}, \frac{16}{729})$ .

$(0, 0), (\frac{2}{9}, \frac{16}{729})$

Step 10