Calculus Examples

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

Step 1

Write $y = {(x - 2)}^{3} + 3$ as a function.

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

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

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

Step 2.1.2

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

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Step 2.1.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = x - 2$ .

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Step 2.1.2.1.1

To apply the Chain Rule, set $u$ as $x - 2$ .

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

Step 2.1.2.1.2

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

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

Step 2.1.2.1.3

Replace all occurrences of $u$ with $x - 2$ .

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

Step 2.1.2.2

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

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

Step 2.1.2.3

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

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

Step 2.1.2.4

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

$f' (x) = 3 {(x - 2)}^{2} (1 + 0) + \frac{d}{d x} (3)$

Step 2.1.2.5

Add $1$ and $0$ .

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

Step 2.1.2.6

Multiply $3$ by $1$ .

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

Step 2.1.3

Differentiate using the Constant Rule.

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Step 2.1.3.1

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

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

Step 2.1.3.2

Add $3 {(x - 2)}^{2}$ and $0$ .

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

Step 2.2

Find the second derivative.

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Step 2.2.1

Rewrite ${(x - 2)}^{2}$ as $(x - 2) (x - 2)$ .

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

Step 2.2.2

Expand $(x - 2) (x - 2)$ using the FOIL Method.

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Step 2.2.2.1

Apply the distributive property.

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

Step 2.2.2.2

Apply the distributive property.

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

Step 2.2.2.3

Apply the distributive property.

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

Step 2.2.3

Simplify and combine like terms.

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Step 2.2.3.1

Simplify each term.

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Step 2.2.3.1.1

Multiply $x$ by $x$ .

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

Step 2.2.3.1.2

Move $- 2$ to the left of $x$ .

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

Step 2.2.3.1.3

Multiply $- 2$ by $- 2$ .

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

Step 2.2.3.2

Subtract $2 x$ from $- 2 x$ .

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

Step 2.2.4

Since $3$ is constant with respect to $x$ , the derivative of $3 (x^{2} - 4 x + 4)$ with respect to $x$ is $3 \frac{d}{d x} [x^{2} - 4 x + 4]$ .

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Multiply $- 4$ by $1$ .

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

Step 2.2.10

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

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

Step 2.2.11

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

$f'' (x) = 3 (2 x - 4)$

Step 2.2.12

Simplify.

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Step 2.2.12.1

Apply the distributive property.

$f'' (x) = 3 (2 x) + 3 \cdot - 4$

Step 2.2.12.2

Combine terms.

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Step 2.2.12.2.1

Multiply $2$ by $3$ .

$f'' (x) = 6 x + 3 \cdot - 4$

Step 2.2.12.2.2

Multiply $3$ by $- 4$ .

$f'' (x) = 6 x - 12$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $6 x - 12$ .

$6 x - 12$

Step 3

Set the second derivative equal to $0$ then solve the equation $6 x - 12 = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$6 x - 12 = 0$

Step 3.2

Add $12$ to both sides of the equation.

$6 x = 12$

Step 3.3

Divide each term in $6 x = 12$ by $6$ and simplify.

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Step 3.3.1

Divide each term in $6 x = 12$ by $6$ .

$\frac{6 x}{6} = \frac{12}{6}$

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.

$\frac{6 x}{6} = \frac{12}{6}$

Step 3.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{12}{6}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $12$ by $6$ .

$x = 2$

Step 4

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

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Step 4.1

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

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Step 4.1.1

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

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

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

Subtract $2$ from $2$ .

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

Step 4.1.2.1.2

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

$f (2) = 0 + 3$

Step 4.1.2.2

Add $0$ and $3$ .

$f (2) = 3$

Step 4.1.2.3

The final answer is $3$ .

$3$

Step 4.2

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

$(2, 3)$

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.

$f'' (1.9) = 6 (1.9) - 12$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $6$ by $1.9$ .

$f'' (1.9) = 11.4 - 12$

Step 6.2.2

Subtract $12$ from $11.4$ .

$f'' (1.9) = - 0.6$

Step 6.2.3

The final answer is $- 0.6$ .

$- 0.6$

Step 6.3

At $1.9$ , the second derivative is $- 0.6$ . 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.

$f'' (2.1) = 6 (2.1) - 12$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $6$ by $2.1$ .

$f'' (2.1) = 12.6 - 12$

Step 7.2.2

Subtract $12$ from $12.6$ .

$f'' (2.1) = 0.6$

Step 7.2.3

The final answer is $0.6$ .

$0.6$

Step 7.3

At $2.1$ , the second derivative is $0.6$ . 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 $(2, 3)$ .

$(2, 3)$

Step 9