Find the Local Maxima and Minima f(x)=2x^3-6x^2-18x+5

$f (x) = 2 x^{3} - 6 x^{2} - 18 x + 5$

Find the first derivative of the function.

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By the Sum Rule, the derivative of $2 x^{3} - 6 x^{2} - 18 x + 5$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$

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

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

$2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$

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

$2 (3 x^{2}) + \frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$

Multiply $3$ by $2$ .

$6 x^{2} + \frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$

Evaluate $\frac{d}{d x} [- 6 x^{2}]$ .

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Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x^{2}$ with respect to $x$ is $- 6 \frac{d}{d x} [x^{2}]$ .

$6 x^{2} - 6 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$

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

$6 x^{2} - 6 (2 x) + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$

Multiply $2$ by $- 6$ .

$6 x^{2} - 12 x + \frac{d}{d x} [- 18 x] + \frac{d}{d x} [5]$

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

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

$6 x^{2} - 12 x - 18 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

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

$6 x^{2} - 12 x - 18 \cdot 1 + \frac{d}{d x} [5]$

Multiply $- 18$ by $1$ .

$6 x^{2} - 12 x - 18 + \frac{d}{d x} [5]$

Differentiate using the Constant Rule.

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Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$6 x^{2} - 12 x - 18 + 0$

Add $6 x^{2} - 12 x - 18$ and $0$ .

$6 x^{2} - 12 x - 18$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $6 x^{2} - 12 x - 18$ with respect to $x$ is $\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 12 x] + \frac{d}{d x} [- 18]$ .

$f'' (x) = \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (- 12 x) + \frac{d}{d x} (- 18)$

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

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Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 6 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 12 x) + \frac{d}{d x} (- 18)$

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

$f'' (x) = 6 (2 x) + \frac{d}{d x} (- 12 x) + \frac{d}{d x} (- 18)$

Multiply $2$ by $6$ .

$f'' (x) = 12 x + \frac{d}{d x} (- 12 x) + \frac{d}{d x} (- 18)$

Evaluate $\frac{d}{d x} [- 12 x]$ .

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Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x$ with respect to $x$ is $- 12 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x - 12 \frac{d}{d x} x + \frac{d}{d x} (- 18)$

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

$f'' (x) = 12 x - 12 \cdot 1 + \frac{d}{d x} (- 18)$

Multiply $- 12$ by $1$ .

$f'' (x) = 12 x - 12 + \frac{d}{d x} (- 18)$

Differentiate using the Constant Rule.

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Since $- 18$ is constant with respect to $x$ , the derivative of $- 18$ with respect to $x$ is $0$ .

$f'' (x) = 12 x - 12 + 0$

Add $12 x - 12$ and $0$ .

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

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

$6 x^{2} - 12 x - 18 = 0$

Factor the left side of the equation.

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Factor $6$ out of $6 x^{2} - 12 x - 18$ .

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

$6 (x^{2}) - 12 x - 18 = 0$

Factor $6$ out of $- 12 x$ .

$6 (x^{2}) + 6 (- 2 x) - 18 = 0$

Factor $6$ out of $- 18$ .

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

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

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

Factor $6$ out of $6 (x^{2} - 2 x) + 6 \cdot - 3$ .

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

Factor.

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Factor $x^{2} - 2 x - 3$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$6 ((x - 3) (x + 1)) = 0$

Remove unnecessary parentheses.

$6 (x - 3) (x + 1) = 0$

Divide each term by $6$ and simplify.

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Divide each term in $6 (x - 3) (x + 1) = 0$ by $6$ .

$\frac{6 (x - 3) (x + 1)}{6} = \frac{0}{6}$

Simplify $\frac{6 (x - 3) (x + 1)}{6}$ .

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Cancel the common factor of $6$ .

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Cancel the common factor.

$\frac{6 (x - 3) (x + 1)}{6} = \frac{0}{6}$

Divide $(x - 3) (x + 1)$ by $1$ .

$(x - 3) (x + 1) = \frac{0}{6}$

Expand $(x - 3) (x + 1)$ using the FOIL Method.

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Apply the distributive property.

$x (x + 1) - 3 (x + 1) = \frac{0}{6}$

Apply the distributive property.

$x \cdot x + x \cdot 1 - 3 (x + 1) = \frac{0}{6}$

Apply the distributive property.

$x \cdot x + x \cdot 1 - 3 x - 3 \cdot 1 = \frac{0}{6}$

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ .

$x^{2} + x \cdot 1 - 3 x - 3 \cdot 1 = \frac{0}{6}$

Multiply $x$ by $1$ .

$x^{2} + x - 3 x - 3 \cdot 1 = \frac{0}{6}$

Multiply $- 3$ by $1$ .

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

Subtract $3 x$ from $x$ .

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

Divide $0$ by $6$ .

$x^{2} - 2 x - 3 = 0$

Factor $x^{2} - 2 x - 3$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Evaluate the second derivative at $x = 3$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 (3) - 12$

Evaluate the second derivative.

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Multiply $12$ by $3$ .

$36 - 12$

Subtract $12$ from $36$ .

$24$

$x = 3$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 3$ is a local minimum

Find the y-value when $x = 3$ .

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

$f (3) = 2 {(3)}^{3} - 6 {(3)}^{2} - 18 \cdot 3 + 5$

Simplify the result.

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Simplify each term.

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

$f (3) = 2 \cdot 27 - 6 {(3)}^{2} - 18 \cdot 3 + 5$

Multiply $2$ by $27$ .

$f (3) = 54 - 6 {(3)}^{2} - 18 \cdot 3 + 5$

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

$f (3) = 54 - 6 \cdot 9 - 18 \cdot 3 + 5$

Multiply $- 6$ by $9$ .

$f (3) = 54 - 54 - 18 \cdot 3 + 5$

Multiply $- 18$ by $3$ .

$f (3) = 54 - 54 - 54 + 5$

Simplify by adding and subtracting.

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Subtract $54$ from $54$ .

$f (3) = 0 - 54 + 5$

Subtract $54$ from $0$ .

$f (3) = - 54 + 5$

Add $- 54$ and $5$ .

$f (3) = - 49$

The final answer is $- 49$ .

$y = - 49$

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 (- 1) - 12$

Evaluate the second derivative.

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Multiply $12$ by $- 1$ .

$- 12 - 12$

Subtract $12$ from $- 12$ .

$- 24$

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

Find the y-value when $x = - 1$ .

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Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = 2 {(- 1)}^{3} - 6 {(- 1)}^{2} - 18 \cdot - 1 + 5$

Simplify the result.

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Simplify each term.

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Raise $- 1$ to the power of $3$ .

$f (- 1) = 2 \cdot - 1 - 6 {(- 1)}^{2} - 18 \cdot - 1 + 5$

Multiply $2$ by $- 1$ .

$f (- 1) = - 2 - 6 {(- 1)}^{2} - 18 \cdot - 1 + 5$

Raise $- 1$ to the power of $2$ .

$f (- 1) = - 2 - 6 \cdot 1 - 18 \cdot - 1 + 5$

Multiply $- 6$ by $1$ .

$f (- 1) = - 2 - 6 - 18 \cdot - 1 + 5$

Multiply $- 18$ by $- 1$ .

$f (- 1) = - 2 - 6 + 18 + 5$

Simplify by adding and subtracting.

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Subtract $6$ from $- 2$ .

$f (- 1) = - 8 + 18 + 5$

Add $- 8$ and $18$ .

$f (- 1) = 10 + 5$

Add $10$ and $5$ .

$f (- 1) = 15$

The final answer is $15$ .

$y = 15$

These are the local extrema for $f (x) = 2 x^{3} - 6 x^{2} - 18 x + 5$ .

$(3, - 49)$ is a local minima

$(- 1, 15)$ is a local maxima

Find the Local Maxima and Minima f(x)=2x^3-6x^2-18x+5

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