Calculus Examples

$f (x, y) = {(x - 1)}^{2} + y^{3} - 3 y^{2} - 9 y + 5$

Step 1

Write $f (x, y) - x^{2} + 2 x - y^{3} + 3 y^{2} + 9 y - 6 = 0$ as a function.

$f (x) = f (x, y) - x^{2} + 2 x - y^{3} + 3 y^{2} + 9 y - 6$

Step 2

Move all the expressions to the left side of the equation.

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Step 2.1

Subtract ${(x - 1)}^{2}$ from both sides of the equation.

$f (x, y) - {(x - 1)}^{2} = y^{3} - 3 y^{2} - 9 y + 5$

Step 2.2

Subtract $y^{3}$ from both sides of the equation.

$f (x, y) - {(x - 1)}^{2} - y^{3} = - 3 y^{2} - 9 y + 5$

Step 2.3

Add $3 y^{2}$ to both sides of the equation.

$f (x, y) - {(x - 1)}^{2} - y^{3} + 3 y^{2} = - 9 y + 5$

Step 2.4

Add $9 y$ to both sides of the equation.

$f (x, y) - {(x - 1)}^{2} - y^{3} + 3 y^{2} + 9 y = 5$

Step 2.5

Subtract $5$ from both sides of the equation.

$f (x, y) - {(x - 1)}^{2} - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3

Simplify $f (x, y) - {(x - 1)}^{2} - y^{3} + 3 y^{2} + 9 y - 5$ .

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Step 3.1

Simplify each term.

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Step 3.1.1

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$f (x, y) - ((x - 1) (x - 1)) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 3.1.2.1

Apply the distributive property.

$f (x, y) - (x (x - 1) - 1 (x - 1)) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.2.2

Apply the distributive property.

$f (x, y) - (x \cdot x + x \cdot - 1 - 1 (x - 1)) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.2.3

Apply the distributive property.

$f (x, y) - (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.3

Simplify and combine like terms.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

Multiply $x$ by $x$ .

$f (x, y) - (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.3.1.2

Move $- 1$ to the left of $x$ .

$f (x, y) - (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$f (x, y) - (x^{2} - x - 1 x - 1 \cdot - 1) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.3.1.4

Rewrite $- 1 x$ as $- x$ .

$f (x, y) - (x^{2} - x - x - 1 \cdot - 1) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.3.1.5

Multiply $- 1$ by $- 1$ .

$f (x, y) - (x^{2} - x - x + 1) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.3.2

Subtract $x$ from $- x$ .

$f (x, y) - (x^{2} - 2 x + 1) - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.4

Apply the distributive property.

$f (x, y) - x^{2} - (- 2 x) - 1 \cdot 1 - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.5

Simplify.

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Step 3.1.5.1

Multiply $- 2$ by $- 1$ .

$f (x, y) - x^{2} + 2 x - 1 \cdot 1 - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.1.5.2

Multiply $- 1$ by $1$ .

$f (x, y) - x^{2} + 2 x - 1 - y^{3} + 3 y^{2} + 9 y - 5 = 0$

Step 3.2

Subtract $5$ from $- 1$ .

$f (x, y) - x^{2} + 2 x - y^{3} + 3 y^{2} + 9 y - 6 = 0$

Step 4

Find the first derivative of the function.

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Step 4.1

By the Sum Rule, the derivative of $f (x, y) - x^{2} + 2 x - y^{3} + 3 y^{2} + 9 y - 6$ with respect to $x$ is $\frac{d}{d x} [f (x, y)] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$ .

$\frac{d}{d x} [f (x, y)] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.2

Multiply $f$ by each element of the matrix.

$\frac{d}{d x} [(f x, f y)] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.3

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

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Step 4.3.1

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

$\frac{d}{d x} [(f x, f y)] - \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.3.2

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

$\frac{d}{d x} [(f x, f y)] - (2 x) + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.3.3

Multiply $2$ by $- 1$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.4

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

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Step 4.4.1

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

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 \frac{d}{d x} [x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.4.2

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

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 \cdot 1 + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.4.3

Multiply $2$ by $1$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.5

Differentiate using the Constant Rule.

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Step 4.5.1

Since $- y^{3}$ is constant with respect to $x$ , the derivative of $- y^{3}$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.5.2

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 y^{2}$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 4.5.3

Since $9 y$ is constant with respect to $x$ , the derivative of $9 y$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + 0 + \frac{d}{d x} [- 6]$

Step 4.5.4

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + 0 + 0$

Step 4.6

Simplify.

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Step 4.6.1

Combine terms.

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Step 4.6.1.1

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + 0$

Step 4.6.1.2

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0$

Step 4.6.1.3

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0$

Step 4.6.1.4

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2$

Step 4.6.2

Reorder terms.

$- 2 x + \frac{d}{d x} [(f x, f y)] + 2$

Step 5

Find the second derivative of the function.

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Step 5.1

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

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

Multiply $- 2$ by $1$ .

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

Step 5.3

Evaluate $\frac{d}{d x} [\frac{d}{d x} ((f x, f y))]$ .

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Step 5.3.1

Cancel the common factor of $d$ .

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Step 5.3.1.1

Cancel the common factor.

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

Step 5.3.1.2

Rewrite the expression.

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

Step 5.3.2

Multiply $\frac{1}{x}$ by each element of the matrix.

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

Step 5.3.3

Simplify each element in the matrix.

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Step 5.3.3.1

Cancel the common factor of $x$ .

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Step 5.3.3.1.1

Factor $x$ out of $f x$ .

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

Step 5.3.3.1.2

Cancel the common factor.

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

Step 5.3.3.1.3

Rewrite the expression.

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

Step 5.3.3.2

Multiply $\frac{1}{x} (f y)$ .

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Step 5.3.3.2.1

Combine $f$ and $\frac{1}{x}$ .

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

Step 5.3.3.2.2

Combine $\frac{f}{x}$ and $y$ .

$f'' (x) = - 2 + \frac{d}{d x} ((f, \frac{f y}{x})) + \frac{d}{d x} (2)$

Step 5.4

Differentiate using the Constant Rule.

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Step 5.4.1

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

$f'' (x) = - 2 + \frac{d}{d x} ((f, \frac{f y}{x})) + 0$

Step 5.4.2

Add $- 2 + \frac{d}{d x} [(f, \frac{f y}{x})]$ and $0$ .

$f'' (x) = - 2 + \frac{d}{d x} ((f, \frac{f y}{x}))$

Step 6

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- 2 x + \frac{d}{d x} ((f x, f y)) + 2 = 0$

Step 7

Find the first derivative.

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Step 7.1

Find the first derivative.

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Step 7.1.1

$\frac{d}{d x} [f (x, y)] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.2

Multiply $f$ by each element of the matrix.

$\frac{d}{d x} [(f x, f y)] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.3

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

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Step 7.1.3.1

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

$\frac{d}{d x} [(f x, f y)] - \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.3.2

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

$\frac{d}{d x} [(f x, f y)] - (2 x) + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.3.3

Multiply $2$ by $- 1$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + \frac{d}{d x} [2 x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.4

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

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Step 7.1.4.1

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

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 \frac{d}{d x} [x] + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.4.2

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

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 \cdot 1 + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.4.3

Multiply $2$ by $1$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.5

Differentiate using the Constant Rule.

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Step 7.1.5.1

Since $- y^{3}$ is constant with respect to $x$ , the derivative of $- y^{3}$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + \frac{d}{d x} [3 y^{2}] + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.5.2

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 y^{2}$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + \frac{d}{d x} [9 y] + \frac{d}{d x} [- 6]$

Step 7.1.5.3

Since $9 y$ is constant with respect to $x$ , the derivative of $9 y$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + 0 + \frac{d}{d x} [- 6]$

Step 7.1.5.4

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + 0 + 0$

Step 7.1.6

Simplify.

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Step 7.1.6.1

Combine terms.

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Step 7.1.6.1.1

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0 + 0$

Step 7.1.6.1.2

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0 + 0$

Step 7.1.6.1.3

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2 + 0$

Step 7.1.6.1.4

Add $\frac{d}{d x} [(f x, f y)] - 2 x + 2$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 2 x + 2$

Step 7.1.6.2

Reorder terms.

$f' (x) = - 2 x + \frac{d}{d x} [(f x, f y)] + 2$

Step 7.2

The first derivative of $f (x)$ with respect to $x$ is $- 2 x + \frac{d}{d x} ((f x, f y)) + 2$ .

$- 2 x + \frac{d}{d x} ((f x, f y)) + 2$

Step 8

Set the first derivative equal to $0$ .

$- 2 x + \frac{d}{d x} ((f x, f y)) + 2 = 0$

Step 9

Find the values where the derivative is undefined.

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Step 9.1

Set the denominator in $\frac{d}{d x}$ equal to $0$ to find where the expression is undefined.

$d x = 0$

Step 9.2

Divide each term in $d x = 0$ by $d$ and simplify.

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Step 9.2.1

Divide each term in $d x = 0$ by $d$ .

$\frac{d x}{d} = \frac{0}{d}$

Step 9.2.2

Simplify the left side.

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Step 9.2.2.1

Cancel the common factor of $d$ .

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Step 9.2.2.1.1

Cancel the common factor.

$\frac{d x}{d} = \frac{0}{d}$

Step 9.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{d}$

Step 9.2.3

Simplify the right side.

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Step 9.2.3.1

Divide $0$ by $d$ .

$x = 0$

Step 10

Critical points to evaluate.

$x = 0$

Step 11

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 + \frac{d}{d (0)} ((f, \frac{f y}{0}))$

Step 12

Evaluate the second derivative.

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Step 12.1

Multiply $d$ by $0$ .

$- 2 + \frac{d}{0} (f, \frac{f y}{0})$

Step 12.2

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 13

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Step 14