Calculus Examples

$y = \frac{1}{3} x^{3} - 2 x^{2} + 3 x + 2$

Step 1

Write $y = \frac{1}{3} x^{3} - 2 x^{2} + 3 x + 2$ as a function.

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

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

$f' (x) = \frac{d}{d x} (\frac{1}{3} x^{3}) + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.2

Evaluate $\frac{d}{d x} [\frac{1}{3} x^{3}]$ .

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Step 2.1.2.1

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{1}{3} x^{3}$ with respect to $x$ is $\frac{1}{3} \frac{d}{d x} [x^{3}]$ .

$f' (x) = \frac{1}{3} \cdot \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

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 $n = 3$ .

$f' (x) = \frac{1}{3} \cdot (3 x^{2}) + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.2.3

Combine $3$ and $\frac{1}{3}$ .

$f' (x) = \frac{3}{3} x^{2} + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.2.4

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

$f' (x) = \frac{3 x^{2}}{3} + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.2.5

Cancel the common factor of $3$ .

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Step 2.1.2.5.1

Cancel the common factor.

$f' (x) = \frac{3 x^{2}}{3} + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.2.5.2

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

$f' (x) = x^{2} + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.3

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

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Step 2.1.3.1

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

$f' (x) = x^{2} - 2 \frac{d}{d x} x^{2} + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

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 $n = 2$ .

$f' (x) = x^{2} - 2 (2 x) + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.3.3

Multiply $2$ by $- 2$ .

$f' (x) = x^{2} - 4 x + \frac{d}{d x} (3 x) + \frac{d}{d x} (2)$

Step 2.1.4

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

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Step 2.1.4.1

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

$f' (x) = x^{2} - 4 x + 3 \frac{d}{d x} (x) + \frac{d}{d x} (2)$

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 $n = 1$ .

$f' (x) = x^{2} - 4 x + 3 \cdot 1 + \frac{d}{d x} (2)$

Step 2.1.4.3

Multiply $3$ by $1$ .

$f' (x) = x^{2} - 4 x + 3 + \frac{d}{d x} (2)$

Step 2.1.5

Differentiate using the Constant Rule.

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Step 2.1.5.1

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$f' (x) = x^{2} - 4 x + 3 + 0$

Step 2.1.5.2

Add $x^{2} - 4 x + 3$ and $0$ .

$f' (x) = x^{2} - 4 x + 3$

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate.

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Step 2.2.1.1

By the Sum Rule, the derivative of $x^{2} - 4 x + 3$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [3]$ .

$f'' (x) = \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (3)$

Step 2.2.1.2

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

$f'' (x) = 2 x + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (3)$

Step 2.2.2

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

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Step 2.2.2.1

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

$f'' (x) = 2 x - 4 \frac{d}{d x} x + \frac{d}{d x} (3)$

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 $n = 1$ .

$f'' (x) = 2 x - 4 \cdot 1 + \frac{d}{d x} (3)$

Step 2.2.2.3

Multiply $- 4$ by $1$ .

$f'' (x) = 2 x - 4 + \frac{d}{d x} (3)$

Step 2.2.3

Differentiate using the Constant Rule.

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Step 2.2.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$f'' (x) = 2 x - 4 + 0$

Step 2.2.3.2

Add $2 x - 4$ and $0$ .

$f'' (x) = 2 x - 4$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $2 x - 4$ .

$2 x - 4$

Step 3

Set the second derivative equal to $0$ then solve the equation $2 x - 4 = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$2 x - 4 = 0$

Step 3.2

Add $4$ to both sides of the equation.

$2 x = 4$

Step 3.3

Divide each term in $2 x = 4$ by $2$ and simplify.

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Step 3.3.1

Divide each term in $2 x = 4$ by $2$ .

$\frac{2 x}{2} = \frac{4}{2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{4}{2}$

Step 3.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{4}{2}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $4$ by $2$ .

$x = 2$

Step 4

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

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Step 4.1

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

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Step 4.1.1

Replace the variable $x$ with $2$ in the expression.

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

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 $2$ to the power of $3$ .

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

Step 4.1.2.1.2

Combine $\frac{1}{3}$ and $8$ .

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

Step 4.1.2.1.3

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

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

Step 4.1.2.1.4

Multiply $- 2$ by $4$ .

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

Step 4.1.2.1.5

Multiply $3$ by $2$ .

$f (2) = \frac{8}{3} - 8 + 6 + 2$

Step 4.1.2.2

Find the common denominator.

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Step 4.1.2.2.1

Write $- 8$ as a fraction with denominator $1$ .

$f (2) = \frac{8}{3} + \frac{- 8}{1} + 6 + 2$

Step 4.1.2.2.2

Multiply $\frac{- 8}{1}$ by $\frac{3}{3}$ .

$f (2) = \frac{8}{3} + \frac{- 8}{1} \cdot \frac{3}{3} + 6 + 2$

Step 4.1.2.2.3

Multiply $\frac{- 8}{1}$ by $\frac{3}{3}$ .

$f (2) = \frac{8}{3} + \frac{- 8 \cdot 3}{3} + 6 + 2$

Step 4.1.2.2.4

Write $6$ as a fraction with denominator $1$ .

$f (2) = \frac{8}{3} + \frac{- 8 \cdot 3}{3} + \frac{6}{1} + 2$

Step 4.1.2.2.5

Multiply $\frac{6}{1}$ by $\frac{3}{3}$ .

$f (2) = \frac{8}{3} + \frac{- 8 \cdot 3}{3} + \frac{6}{1} \cdot \frac{3}{3} + 2$

Step 4.1.2.2.6

Multiply $\frac{6}{1}$ by $\frac{3}{3}$ .

$f (2) = \frac{8}{3} + \frac{- 8 \cdot 3}{3} + \frac{6 \cdot 3}{3} + 2$

Step 4.1.2.2.7

Write $2$ as a fraction with denominator $1$ .

$f (2) = \frac{8}{3} + \frac{- 8 \cdot 3}{3} + \frac{6 \cdot 3}{3} + \frac{2}{1}$

Step 4.1.2.2.8

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

$f (2) = \frac{8}{3} + \frac{- 8 \cdot 3}{3} + \frac{6 \cdot 3}{3} + \frac{2}{1} \cdot \frac{3}{3}$

Step 4.1.2.2.9

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

$f (2) = \frac{8}{3} + \frac{- 8 \cdot 3}{3} + \frac{6 \cdot 3}{3} + \frac{2 \cdot 3}{3}$

Step 4.1.2.3

Combine the numerators over the common denominator.

$f (2) = \frac{8 - 8 \cdot 3 + 6 \cdot 3 + 2 \cdot 3}{3}$

Step 4.1.2.4

Simplify each term.

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Step 4.1.2.4.1

Multiply $- 8$ by $3$ .

$f (2) = \frac{8 - 24 + 6 \cdot 3 + 2 \cdot 3}{3}$

Step 4.1.2.4.2

Multiply $6$ by $3$ .

$f (2) = \frac{8 - 24 + 18 + 2 \cdot 3}{3}$

Step 4.1.2.4.3

Multiply $2$ by $3$ .

$f (2) = \frac{8 - 24 + 18 + 6}{3}$

Step 4.1.2.5

Simplify by adding and subtracting.

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Step 4.1.2.5.1

Subtract $24$ from $8$ .

$f (2) = \frac{- 16 + 18 + 6}{3}$

Step 4.1.2.5.2

Add $- 16$ and $18$ .

$f (2) = \frac{2 + 6}{3}$

Step 4.1.2.5.3

Add $2$ and $6$ .

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

Step 4.1.2.6

The final answer is $\frac{8}{3}$ .

$\frac{8}{3}$

Step 4.2

The point found by substituting $2$ in $f (x) = \frac{1}{3} x^{3} - 2 x^{2} + 3 x + 2$ is $(2, \frac{8}{3})$ . This point can be an inflection point.

$(2, \frac{8}{3})$

Step 5

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

$(- \infty, 2) \cup (2, \infty)$

Step 6

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

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Step 6.1

Replace the variable $x$ with $1.9$ in the expression.

$f'' (1.9) = 2 (1.9) - 4$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $1.9$ .

$f'' (1.9) = 3.8 - 4$

Step 6.2.2

Subtract $4$ from $3.8$ .

$f'' (1.9) = - 0.2$

Step 6.2.3

The final answer is $- 0.2$ .

$- 0.2$

Step 6.3

At $1.9$ , the second derivative is $- 0.2$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 2)$

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

Step 7

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

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Step 7.1

Replace the variable $x$ with $2.1$ in the expression.

$f'' (2.1) = 2 (2.1) - 4$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $2.1$ .

$f'' (2.1) = 4.2 - 4$

Step 7.2.2

Subtract $4$ from $4.2$ .

$f'' (2.1) = 0.2$

Step 7.2.3

The final answer is $0.2$ .

$0.2$

Step 7.3

At $2.1$ , the second derivative is $0.2$ . Since this is positive, the second derivative is increasing on the interval $(2, \infty)$ .

Increasing on $(2, \infty)$ 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. The inflection point in this case is $(2, \frac{8}{3})$ .

$(2, \frac{8}{3})$

Step 9