Calculus Examples

$f (x) = x^{3} - \frac{3}{2} x^{2} - 36 x$

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.

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Step 1.1.1.1

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

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

Step 1.1.1.2

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

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

Step 1.1.2

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

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Step 1.1.2.1

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

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

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 $n = 2$ .

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

Step 1.1.2.3

Multiply $2$ by $- 1$ .

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

Step 1.1.2.4

Combine $- 2$ and $\frac{3}{2}$ .

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

Step 1.1.2.5

Multiply $- 2$ by $3$ .

$3 x^{2} + \frac{- 6}{2} x + \frac{d}{d x} [- 36 x]$

Step 1.1.2.6

Combine $\frac{- 6}{2}$ and $x$ .

$3 x^{2} + \frac{- 6 x}{2} + \frac{d}{d x} [- 36 x]$

Step 1.1.2.7

Cancel the common factor of $- 6$ and $2$ .

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Step 1.1.2.7.1

Factor $2$ out of $- 6 x$ .

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

Step 1.1.2.7.2

Cancel the common factors.

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Step 1.1.2.7.2.1

Factor $2$ out of $2$ .

$3 x^{2} + \frac{2 (- 3 x)}{2 (1)} + \frac{d}{d x} [- 36 x]$

Step 1.1.2.7.2.2

Cancel the common factor.

$3 x^{2} + \frac{2 (- 3 x)}{2 \cdot 1} + \frac{d}{d x} [- 36 x]$

Step 1.1.2.7.2.3

Rewrite the expression.

$3 x^{2} + \frac{- 3 x}{1} + \frac{d}{d x} [- 36 x]$

Step 1.1.2.7.2.4

Divide $- 3 x$ by $1$ .

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

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 $n = 1$ .

$3 x^{2} - 3 x - 36 \cdot 1$

Step 1.1.3.3

Multiply $- 36$ by $1$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

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

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

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 $n = 2$ .

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

Step 1.2.2.3

Multiply $2$ by $3$ .

$6 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [- 36]$

Step 1.2.3

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

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Step 1.2.3.1

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

$6 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [- 36]$

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 $n = 1$ .

$6 x - 3 \cdot 1 + \frac{d}{d x} [- 36]$

Step 1.2.3.3

Multiply $- 3$ by $1$ .

$6 x - 3 + \frac{d}{d x} [- 36]$

Step 1.2.4

Differentiate using the Constant Rule.

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Step 1.2.4.1

Since $- 36$ is constant with respect to $x$ , the derivative of $- 36$ with respect to $x$ is $0$ .

$6 x - 3 + 0$

Step 1.2.4.2

Add $6 x - 3$ and $0$ .

$f'' (x) = 6 x - 3$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $6 x - 3$ .

$6 x - 3$

Step 2

Set the second derivative equal to $0$ then solve the equation $6 x - 3 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$6 x - 3 = 0$

Step 2.2

Add $3$ to both sides of the equation.

$6 x = 3$

Step 2.3

Divide each term in $6 x = 3$ by $6$ and simplify.

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Step 2.3.1

Divide each term in $6 x = 3$ by $6$ .

$\frac{6 x}{6} = \frac{3}{6}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $6$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{3}{6}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{6}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $3$ and $6$ .

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Step 2.3.3.1.1

Factor $3$ out of $3$ .

$x = \frac{3 (1)}{6}$

Step 2.3.3.1.2

Cancel the common factors.

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Step 2.3.3.1.2.1

Factor $3$ out of $6$ .

$x = \frac{3 \cdot 1}{3 \cdot 2}$

Step 2.3.3.1.2.2

Cancel the common factor.

$x = \frac{3 \cdot 1}{3 \cdot 2}$

Step 2.3.3.1.2.3

Rewrite the expression.

$x = \frac{1}{2}$

Step 3

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

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Step 3.1

Substitute $\frac{1}{2}$ in $f (x) = x^{3} - \frac{3}{2} x^{2} - 36 x$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = {(\frac{1}{2})}^{3} - \frac{3}{2} \cdot {(\frac{1}{2})}^{2} - 36 (\frac{1}{2})$

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 $\frac{1}{2}$ .

$f (\frac{1}{2}) = \frac{1^{3}}{2^{3}} - \frac{3}{2} \cdot {(\frac{1}{2})}^{2} - 36 (\frac{1}{2})$

Step 3.1.2.1.2

One to any power is one.

$f (\frac{1}{2}) = \frac{1}{2^{3}} - \frac{3}{2} \cdot {(\frac{1}{2})}^{2} - 36 (\frac{1}{2})$

Step 3.1.2.1.3

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

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{2} \cdot {(\frac{1}{2})}^{2} - 36 (\frac{1}{2})$

Step 3.1.2.1.4

Apply the product rule to $\frac{1}{2}$ .

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{2} \cdot \frac{1^{2}}{2^{2}} - 36 (\frac{1}{2})$

Step 3.1.2.1.5

One to any power is one.

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{2} \cdot \frac{1}{2^{2}} - 36 (\frac{1}{2})$

Step 3.1.2.1.6

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

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{2} \cdot \frac{1}{4} - 36 (\frac{1}{2})$

Step 3.1.2.1.7

Multiply $- \frac{3}{2} \cdot \frac{1}{4}$ .

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Step 3.1.2.1.7.1

Multiply $\frac{1}{4}$ by $\frac{3}{2}$ .

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{4 \cdot 2} - 36 (\frac{1}{2})$

Step 3.1.2.1.7.2

Multiply $4$ by $2$ .

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{8} - 36 (\frac{1}{2})$

Step 3.1.2.1.8

Cancel the common factor of $2$ .

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Step 3.1.2.1.8.1

Factor $2$ out of $- 36$ .

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{8} + 2 (- 18) (\frac{1}{2})$

Step 3.1.2.1.8.2

Cancel the common factor.

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{8} + 2 \cdot (- 18 (\frac{1}{2}))$

Step 3.1.2.1.8.3

Rewrite the expression.

$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{8} - 18$

Step 3.1.2.2

Combine fractions.

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Step 3.1.2.2.1

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = - 18 + \frac{1 - 3}{8}$

Step 3.1.2.2.2

Subtract $3$ from $1$ .

$f (\frac{1}{2}) = - 18 + \frac{- 2}{8}$

Step 3.1.2.3

Simplify each term.

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Step 3.1.2.3.1

Cancel the common factor of $- 2$ and $8$ .

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Step 3.1.2.3.1.1

Factor $2$ out of $- 2$ .

$f (\frac{1}{2}) = - 18 + \frac{2 (- 1)}{8}$

Step 3.1.2.3.1.2

Cancel the common factors.

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Step 3.1.2.3.1.2.1

Factor $2$ out of $8$ .

$f (\frac{1}{2}) = - 18 + \frac{2 \cdot - 1}{2 \cdot 4}$

Step 3.1.2.3.1.2.2

Cancel the common factor.

$f (\frac{1}{2}) = - 18 + \frac{2 \cdot - 1}{2 \cdot 4}$

Step 3.1.2.3.1.2.3

Rewrite the expression.

$f (\frac{1}{2}) = - 18 + \frac{- 1}{4}$

Step 3.1.2.3.2

Move the negative in front of the fraction.

$f (\frac{1}{2}) = - 18 - \frac{1}{4}$

Step 3.1.2.4

To write $- 18$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (\frac{1}{2}) = - 18 \cdot \frac{4}{4} - \frac{1}{4}$

Step 3.1.2.5

Combine $- 18$ and $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{- 18 \cdot 4}{4} - \frac{1}{4}$

Step 3.1.2.6

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{- 18 \cdot 4 - 1}{4}$

Step 3.1.2.7

Simplify the numerator.

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Step 3.1.2.7.1

Multiply $- 18$ by $4$ .

$f (\frac{1}{2}) = \frac{- 72 - 1}{4}$

Step 3.1.2.7.2

Subtract $1$ from $- 72$ .

$f (\frac{1}{2}) = \frac{- 73}{4}$

Step 3.1.2.8

Move the negative in front of the fraction.

$f (\frac{1}{2}) = - \frac{73}{4}$

Step 3.1.2.9

The final answer is $- \frac{73}{4}$ .

$- \frac{73}{4}$

Step 3.2

The point found by substituting $\frac{1}{2}$ in $f (x) = x^{3} - \frac{3}{2} x^{2} - 36 x$ is $(\frac{1}{2}, - \frac{73}{4})$ . This point can be an inflection point.

$(\frac{1}{2}, - \frac{73}{4})$

Step 4

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

$(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $0.4$ in the expression.

$f'' (0.4) = 6 (0.4) - 3$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $6$ by $0.4$ .

$f'' (0.4) = 2.4 - 3$

Step 5.2.2

Subtract $3$ from $2.4$ .

$f'' (0.4) = - 0.6$

Step 5.2.3

The final answer is $- 0.6$ .

$- 0.6$

Step 5.3

At $0.4$ , the second derivative is $- 0.6$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{1}{2})$

Decreasing on $(- \infty, \frac{1}{2})$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(\frac{1}{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.6$ in the expression.

$f'' (0.6) = 6 (0.6) - 3$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $6$ by $0.6$ .

$f'' (0.6) = 3.6 - 3$

Step 6.2.2

Subtract $3$ from $3.6$ .

$f'' (0.6) = 0.6$

Step 6.2.3

The final answer is $0.6$ .

$0.6$

Step 6.3

At $0.6$ , the second derivative is $0.6$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{2}, \infty)$ .

Increasing on $(\frac{1}{2}, \infty)$ 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 $(\frac{1}{2}, - \frac{73}{4})$ .

$(\frac{1}{2}, - \frac{73}{4})$

Step 8