Calculus Examples

$- \frac{1}{2} x^{4} - 6 x^{3} - 27 x^{2}$

Step 1

Write $- \frac{1}{2} x^{4} - 6 x^{3} - 27 x^{2}$ as a function.

$f (x) = - \frac{1}{2} x^{4} - 6 x^{3} - 27 x^{2}$

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 $- \frac{1}{2} x^{4} - 6 x^{3} - 27 x^{2}$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{2} x^{4}] + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$ .

$\frac{d}{d x} [- \frac{1}{2} x^{4}] + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2

Evaluate $\frac{d}{d x} [- \frac{1}{2} x^{4}]$ .

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Step 2.1.2.1

Since $- \frac{1}{2}$ is constant with respect to $x$ , the derivative of $- \frac{1}{2} x^{4}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [x^{4}]$ .

$- \frac{1}{2} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

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 $n = 4$ .

$- \frac{1}{2} (4 x^{3}) + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.3

Multiply $4$ by $- 1$ .

$- 4 (\frac{1}{2}) x^{3} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.4

Combine $- 4$ and $\frac{1}{2}$ .

$\frac{- 4}{2} x^{3} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.5

Combine $\frac{- 4}{2}$ and $x^{3}$ .

$\frac{- 4 x^{3}}{2} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.6

Cancel the common factor of $- 4$ and $2$ .

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Step 2.1.2.6.1

Factor $2$ out of $- 4 x^{3}$ .

$\frac{2 (- 2 x^{3})}{2} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.6.2

Cancel the common factors.

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Step 2.1.2.6.2.1

Factor $2$ out of $2$ .

$\frac{2 (- 2 x^{3})}{2 (1)} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.6.2.2

Cancel the common factor.

$\frac{2 (- 2 x^{3})}{2 \cdot 1} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.6.2.3

Rewrite the expression.

$\frac{- 2 x^{3}}{1} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.2.6.2.4

Divide $- 2 x^{3}$ by $1$ .

$- 2 x^{3} + \frac{d}{d x} [- 6 x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- 6 x^{3}]$ .

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Step 2.1.3.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x^{3}$ with respect to $x$ is $- 6 \frac{d}{d x} [x^{3}]$ .

$- 2 x^{3} - 6 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 27 x^{2}]$

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 $n = 3$ .

$- 2 x^{3} - 6 (3 x^{2}) + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.3.3

Multiply $3$ by $- 6$ .

$- 2 x^{3} - 18 x^{2} + \frac{d}{d x} [- 27 x^{2}]$

Step 2.1.4

Evaluate $\frac{d}{d x} [- 27 x^{2}]$ .

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Step 2.1.4.1

Since $- 27$ is constant with respect to $x$ , the derivative of $- 27 x^{2}$ with respect to $x$ is $- 27 \frac{d}{d x} [x^{2}]$ .

$- 2 x^{3} - 18 x^{2} - 27 \frac{d}{d x} [x^{2}]$

Step 2.1.4.2

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

$- 2 x^{3} - 18 x^{2} - 27 (2 x)$

Step 2.1.4.3

Multiply $2$ by $- 27$ .

$f' (x) = - 2 x^{3} - 18 x^{2} - 54 x$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $- 2 x^{3} - 18 x^{2} - 54 x$ with respect to $x$ is $\frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- 18 x^{2}] + \frac{d}{d x} [- 54 x]$ .

$\frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- 18 x^{2}] + \frac{d}{d x} [- 54 x]$

Step 2.2.2

Evaluate $\frac{d}{d x} [- 2 x^{3}]$ .

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Step 2.2.2.1

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

$- 2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 18 x^{2}] + \frac{d}{d x} [- 54 x]$

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 $n = 3$ .

$- 2 (3 x^{2}) + \frac{d}{d x} [- 18 x^{2}] + \frac{d}{d x} [- 54 x]$

Step 2.2.2.3

Multiply $3$ by $- 2$ .

$- 6 x^{2} + \frac{d}{d x} [- 18 x^{2}] + \frac{d}{d x} [- 54 x]$

Step 2.2.3

Evaluate $\frac{d}{d x} [- 18 x^{2}]$ .

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Step 2.2.3.1

Since $- 18$ is constant with respect to $x$ , the derivative of $- 18 x^{2}$ with respect to $x$ is $- 18 \frac{d}{d x} [x^{2}]$ .

$- 6 x^{2} - 18 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 54 x]$

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 $n = 2$ .

$- 6 x^{2} - 18 (2 x) + \frac{d}{d x} [- 54 x]$

Step 2.2.3.3

Multiply $2$ by $- 18$ .

$- 6 x^{2} - 36 x + \frac{d}{d x} [- 54 x]$

Step 2.2.4

Evaluate $\frac{d}{d x} [- 54 x]$ .

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Step 2.2.4.1

Since $- 54$ is constant with respect to $x$ , the derivative of $- 54 x$ with respect to $x$ is $- 54 \frac{d}{d x} [x]$ .

$- 6 x^{2} - 36 x - 54 \frac{d}{d x} [x]$

Step 2.2.4.2

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

$- 6 x^{2} - 36 x - 54 \cdot 1$

Step 2.2.4.3

Multiply $- 54$ by $1$ .

$f'' (x) = - 6 x^{2} - 36 x - 54$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $- 6 x^{2} - 36 x - 54$ .

$- 6 x^{2} - 36 x - 54$

Step 3

Set the second derivative equal to $0$ then solve the equation $- 6 x^{2} - 36 x - 54 = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$- 6 x^{2} - 36 x - 54 = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $- 6$ out of $- 6 x^{2} - 36 x - 54$ .

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Step 3.2.1.1

Factor $- 6$ out of $- 6 x^{2}$ .

$- 6 x^{2} - 36 x - 54 = 0$

Step 3.2.1.2

Factor $- 6$ out of $- 36 x$ .

$- 6 x^{2} - 6 (6 x) - 54 = 0$

Step 3.2.1.3

Factor $- 6$ out of $- 54$ .

$- 6 x^{2} - 6 (6 x) - 6 \cdot 9 = 0$

Step 3.2.1.4

Factor $- 6$ out of $- 6 (x^{2}) - 6 (6 x)$ .

$- 6 (x^{2} + 6 x) - 6 \cdot 9 = 0$

Step 3.2.1.5

Factor $- 6$ out of $- 6 (x^{2} + 6 x) - 6 (9)$ .

$- 6 (x^{2} + 6 x + 9) = 0$

Step 3.2.2

Factor using the perfect square rule.

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Step 3.2.2.1

Rewrite $9$ as $3^{2}$ .

$- 6 (x^{2} + 6 x + 3^{2}) = 0$

Step 3.2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 3.2.2.3

Rewrite the polynomial.

$- 6 (x^{2} + 2 \cdot x \cdot 3 + 3^{2}) = 0$

Step 3.2.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

$- 6 {(x + 3)}^{2} = 0$

Step 3.3

Divide each term in $- 6 {(x + 3)}^{2} = 0$ by $- 6$ and simplify.

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Step 3.3.1

Divide each term in $- 6 {(x + 3)}^{2} = 0$ by $- 6$ .

$\frac{- 6 {(x + 3)}^{2}}{- 6} = \frac{0}{- 6}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $- 6$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 6 {(x + 3)}^{2}}{- 6} = \frac{0}{- 6}$

Step 3.3.2.1.2

Divide ${(x + 3)}^{2}$ by $1$ .

${(x + 3)}^{2} = \frac{0}{- 6}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $0$ by $- 6$ .

${(x + 3)}^{2} = 0$

Step 3.4

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 3.5

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4

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

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Step 4.1

Substitute $- 3$ in $f (x) = - \frac{1}{2} x^{4} - 6 x^{3} - 27 x^{2}$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = - \frac{1}{2} \cdot {(- 3)}^{4} - 6 {(- 3)}^{3} - 27 {(- 3)}^{2}$

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

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

$f (- 3) = - \frac{1}{2} \cdot 81 - 6 {(- 3)}^{3} - 27 {(- 3)}^{2}$

Step 4.1.2.1.2

Multiply $- \frac{1}{2} \cdot 81$ .

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Step 4.1.2.1.2.1

Multiply $81$ by $- 1$ .

$f (- 3) = - 81 (\frac{1}{2}) - 6 {(- 3)}^{3} - 27 {(- 3)}^{2}$

Step 4.1.2.1.2.2

Combine $- 81$ and $\frac{1}{2}$ .

$f (- 3) = \frac{- 81}{2} - 6 {(- 3)}^{3} - 27 {(- 3)}^{2}$

Step 4.1.2.1.3

Move the negative in front of the fraction.

$f (- 3) = - \frac{81}{2} - 6 {(- 3)}^{3} - 27 {(- 3)}^{2}$

Step 4.1.2.1.4

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

$f (- 3) = - \frac{81}{2} - 6 \cdot - 27 - 27 {(- 3)}^{2}$

Step 4.1.2.1.5

Multiply $- 6$ by $- 27$ .

$f (- 3) = - \frac{81}{2} + 162 - 27 {(- 3)}^{2}$

Step 4.1.2.1.6

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

$f (- 3) = - \frac{81}{2} + 162 - 27 \cdot 9$

Step 4.1.2.1.7

Multiply $- 27$ by $9$ .

$f (- 3) = - \frac{81}{2} + 162 - 243$

Step 4.1.2.2

Find the common denominator.

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Step 4.1.2.2.1

Write $162$ as a fraction with denominator $1$ .

$f (- 3) = - \frac{81}{2} + \frac{162}{1} - 243$

Step 4.1.2.2.2

Multiply $\frac{162}{1}$ by $\frac{2}{2}$ .

$f (- 3) = - \frac{81}{2} + \frac{162}{1} \cdot \frac{2}{2} - 243$

Step 4.1.2.2.3

Multiply $\frac{162}{1}$ by $\frac{2}{2}$ .

$f (- 3) = - \frac{81}{2} + \frac{162 \cdot 2}{2} - 243$

Step 4.1.2.2.4

Write $- 243$ as a fraction with denominator $1$ .

$f (- 3) = - \frac{81}{2} + \frac{162 \cdot 2}{2} + \frac{- 243}{1}$

Step 4.1.2.2.5

Multiply $\frac{- 243}{1}$ by $\frac{2}{2}$ .

$f (- 3) = - \frac{81}{2} + \frac{162 \cdot 2}{2} + \frac{- 243}{1} \cdot \frac{2}{2}$

Step 4.1.2.2.6

Multiply $\frac{- 243}{1}$ by $\frac{2}{2}$ .

$f (- 3) = - \frac{81}{2} + \frac{162 \cdot 2}{2} + \frac{- 243 \cdot 2}{2}$

Step 4.1.2.3

Combine the numerators over the common denominator.

$f (- 3) = \frac{- 81 + 162 \cdot 2 - 243 \cdot 2}{2}$

Step 4.1.2.4

Simplify each term.

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Step 4.1.2.4.1

Multiply $162$ by $2$ .

$f (- 3) = \frac{- 81 + 324 - 243 \cdot 2}{2}$

Step 4.1.2.4.2

Multiply $- 243$ by $2$ .

$f (- 3) = \frac{- 81 + 324 - 486}{2}$

Step 4.1.2.5

Simplify the expression.

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Step 4.1.2.5.1

Add $- 81$ and $324$ .

$f (- 3) = \frac{243 - 486}{2}$

Step 4.1.2.5.2

Subtract $486$ from $243$ .

$f (- 3) = \frac{- 243}{2}$

Step 4.1.2.5.3

Move the negative in front of the fraction.

$f (- 3) = - \frac{243}{2}$

Step 4.1.2.6

The final answer is $- \frac{243}{2}$ .

$- \frac{243}{2}$

Step 4.2

The point found by substituting $- 3$ in $f (x) = - \frac{1}{2} x^{4} - 6 x^{3} - 27 x^{2}$ is $(- 3, - \frac{243}{2})$ . This point can be an inflection point.

$(- 3, - \frac{243}{2})$

Step 5

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

$(- \infty, - 3) \cup (- 3, \infty)$

Step 6

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

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Step 6.1

Replace the variable $x$ with $- 3.1$ in the expression.

$f'' (- 3.1) = - 6 {(- 3.1)}^{2} - 36 \cdot - 3.1 - 54$

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 $- 3.1$ to the power of $2$ .

$f'' (- 3.1) = - 6 \cdot 9.61 - 36 \cdot - 3.1 - 54$

Step 6.2.1.2

Multiply $- 6$ by $9.61$ .

$f'' (- 3.1) = - 57.66 - 36 \cdot - 3.1 - 54$

Step 6.2.1.3

Multiply $- 36$ by $- 3.1$ .

$f'' (- 3.1) = - 57.66 + 111.6 - 54$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Add $- 57.66$ and $111.6$ .

$f'' (- 3.1) = 53.94 - 54$

Step 6.2.2.2

Subtract $54$ from $53.94$ .

$f'' (- 3.1) = - 0.06$

Step 6.2.3

The final answer is $- 0.06$ .

$- 0.06$

Step 6.3

At $- 3.1$ , the second derivative is $- 0.06$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 3)$

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

Step 7

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

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Step 7.1

Replace the variable $x$ with $- 2.9$ in the expression.

$f'' (- 2.9) = - 6 {(- 2.9)}^{2} - 36 \cdot - 2.9 - 54$

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 $- 2.9$ to the power of $2$ .

$f'' (- 2.9) = - 6 \cdot 8.41 - 36 \cdot - 2.9 - 54$

Step 7.2.1.2

Multiply $- 6$ by $8.41$ .

$f'' (- 2.9) = - 50.46 - 36 \cdot - 2.9 - 54$

Step 7.2.1.3

Multiply $- 36$ by $- 2.9$ .

$f'' (- 2.9) = - 50.46 + 104.4 - 54$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Add $- 50.46$ and $104.4$ .

$f'' (- 2.9) = 53.94 - 54$

Step 7.2.2.2

Subtract $54$ from $53.94$ .

$f'' (- 2.9) = - 0.06$

Step 7.2.3

The final answer is $- 0.06$ .

$- 0.06$

Step 7.3

At $- 2.9$ , the second derivative is $- 0.06$ . Since this is negative, the second derivative is decreasing on the interval $(- 3, \infty)$

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

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. There are no points on the graph that satisfy these requirements.

No Inflection Points