Find the Local Maxima and Minima g(x)=x^3-2.1x+2

$g (x) = x^{3} - 2.1 x + 2$

Find the first derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $x^{3} - 2.1 x + 2$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 2.1 x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 2.1 x] + \frac{d}{d x} [2]$

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

$3 x^{2} + \frac{d}{d x} [- 2.1 x] + \frac{d}{d x} [2]$

Evaluate $\frac{d}{d x} [- 2.1 x]$ .

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Since $- 2.1$ is constant with respect to $x$ , the derivative of $- 2.1 x$ with respect to $x$ is $- 2.1 \frac{d}{d x} [x]$ .

$3 x^{2} - 2.1 \frac{d}{d x} [x] + \frac{d}{d x} [2]$

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

$3 x^{2} - 2.1 \cdot 1 + \frac{d}{d x} [2]$

Multiply $- 2.1$ by $1$ .

$3 x^{2} - 2.1 + \frac{d}{d x} [2]$

Differentiate using the Constant Rule.

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

$3 x^{2} - 2.1 + 0$

Add $3 x^{2} - 2.1$ and $0$ .

$3 x^{2} - 2.1$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $3 x^{2} - 2.1$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 2.1]$ .

$g'' (x) = \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 2.1)$

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

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

$g'' (x) = 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 2.1)$

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

$g'' (x) = 3 (2 x) + \frac{d}{d x} (- 2.1)$

Multiply $2$ by $3$ .

$g'' (x) = 6 x + \frac{d}{d x} (- 2.1)$

Differentiate using the Constant Rule.

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Since $- 2.1$ is constant with respect to $x$ , the derivative of $- 2.1$ with respect to $x$ is $0$ .

$g'' (x) = 6 x + 0$

Add $6 x$ and $0$ .

$g'' (x) = 6 x$

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

$3 x^{2} - 2.1 = 0$

Add $2.1$ to both sides of the equation.

$3 x^{2} = 2.1$

Divide each term by $3$ and simplify.

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Divide each term in $3 x^{2} = 2.1$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{2.1}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{2.1}{3}$

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

$x^{2} = \frac{2.1}{3}$

Divide $2.1$ by $3$ .

$x^{2} = 0.7$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0.7}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{0.7}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{0.7}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{0.7}, - \sqrt{0.7}$

Evaluate the second derivative at $x = \sqrt{0.7}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (\sqrt{0.7})$

Multiply $6$ by $\sqrt{0.7}$ .

$6 \sqrt{0.7}$

$x = \sqrt{0.7}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \sqrt{0.7}$ is a local minimum

Find the y-value when $x = \sqrt{0.7}$ .

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Replace the variable $x$ with $\sqrt{0.7}$ in the expression.

$g (\sqrt{0.7}) = {(\sqrt{0.7})}^{3} - 2.1 \sqrt{0.7} + 2$

Simplify the result.

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Simplify each term.

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Rewrite ${\sqrt{0.7}}^{3}$ as ${({0.7}^{3})}^{\frac{1}{2}}$ .

$g (\sqrt{0.7}) = \sqrt{{0.7}^{3}} - 2.1 \sqrt{0.7} + 2$

Raise $0.7$ to the power of $3$ .

$g (\sqrt{0.7}) = \sqrt{0.343} - 2.1 \sqrt{0.7} + 2$

Multiply $- 2.1$ by $\sqrt{0.7}$ .

$g (\sqrt{0.7}) = \sqrt{0.343} - 1.75698605 + 2$

Simplify by adding terms.

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Subtract $1.75698605$ from $\sqrt{0.343}$ .

$g (\sqrt{0.7}) = - 1.17132403 + 2$

Add $- 1.17132403$ and $2$ .

$g (\sqrt{0.7}) = 0.82867596$

The final answer is $0.82867596$ .

$y = 0.82867596$

Evaluate the second derivative at $x = - \sqrt{0.7}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (- \sqrt{0.7})$

Multiply $- 1$ by $6$ .

$- 6 \sqrt{0.7}$

$x = - \sqrt{0.7}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - \sqrt{0.7}$ is a local maximum

Find the y-value when $x = - \sqrt{0.7}$ .

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Replace the variable $x$ with $- \sqrt{0.7}$ in the expression.

$g (- \sqrt{0.7}) = {(- \sqrt{0.7})}^{3} - 2.1 (- \sqrt{0.7}) + 2$

Simplify the result.

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Simplify each term.

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Apply the product rule to $- \sqrt{0.7}$ .

$g (- \sqrt{0.7}) = {(- 1)}^{3} {\sqrt{0.7}}^{3} - 2.1 (- \sqrt{0.7}) + 2$

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

$g (- \sqrt{0.7}) = - {\sqrt{0.7}}^{3} - 2.1 (- \sqrt{0.7}) + 2$

Rewrite ${\sqrt{0.7}}^{3}$ as ${({0.7}^{3})}^{\frac{1}{2}}$ .

$g (- \sqrt{0.7}) = - \sqrt{{0.7}^{3}} - 2.1 (- \sqrt{0.7}) + 2$

Raise $0.7$ to the power of $3$ .

$g (- \sqrt{0.7}) = - \sqrt{0.343} - 2.1 (- \sqrt{0.7}) + 2$

Multiply $- 2.1 (- \sqrt{0.7})$ .

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Multiply $- 1$ by $- 2.1$ .

$g (- \sqrt{0.7}) = - \sqrt{0.343} + 2.1 \sqrt{0.7} + 2$

Multiply $2.1$ by $\sqrt{0.7}$ .

$g (- \sqrt{0.7}) = - \sqrt{0.343} + 1.75698605 + 2$

Simplify by adding terms.

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Add $- \sqrt{0.343}$ and $1.75698605$ .

$g (- \sqrt{0.7}) = 1.17132403 + 2$

Add $1.17132403$ and $2$ .

$g (- \sqrt{0.7}) = 3.17132403$

The final answer is $3.17132403$ .

$y = 3.17132403$

These are the local extrema for $g (x) = x^{3} - 2.1 x + 2$ .

$(\sqrt{0.7}, 0.82867596)$ is a local minima

$(- \sqrt{0.7}, 3.17132403)$ is a local maxima

Find the Local Maxima and Minima g(x)=x^3-2.1x+2

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