Find the Maximum/Minimum Value f(x)=x^3-0.9x+2

$f (x) = x^{3} - 0.9 x + 2$

Step 1

Find the first derivative of the function.

Differentiate.

By the Sum Rule, the derivative of $x^{3} - 0.9 x + 2$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 0.9 x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 0.9 x] + \frac{d}{d x} [2]$

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

$3 x^{2} + \frac{d}{d x} [- 0.9 x] + \frac{d}{d x} [2]$

Evaluate $\frac{d}{d x} [- 0.9 x]$ .

Since $- 0.9$ is constant with respect to $x$ , the derivative of $- 0.9 x$ with respect to $x$ is $- 0.9 \frac{d}{d x} [x]$ .

$3 x^{2} - 0.9 \frac{d}{d x} [x] + \frac{d}{d x} [2]$

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

$3 x^{2} - 0.9 \cdot 1 + \frac{d}{d x} [2]$

Multiply $- 0.9$ by $1$ .

$3 x^{2} - 0.9 + \frac{d}{d x} [2]$

Differentiate using the Constant Rule.

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

$3 x^{2} - 0.9 + 0$

Add $3 x^{2} - 0.9$ and $0$ .

$3 x^{2} - 0.9$

Step 2

Find the second derivative of the function.

By the Sum Rule, the derivative of $3 x^{2} - 0.9$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 0.9]$ .

$f'' (x) = \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 0.9)$

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

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

$f'' (x) = 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 0.9)$

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

$f'' (x) = 3 (2 x) + \frac{d}{d x} (- 0.9)$

Multiply $2$ by $3$ .

$f'' (x) = 6 x + \frac{d}{d x} (- 0.9)$

Differentiate using the Constant Rule.

Since $- 0.9$ is constant with respect to $x$ , the derivative of $- 0.9$ with respect to $x$ is $0$ .

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

Add $6 x$ and $0$ .

$f'' (x) = 6 x$

Step 3

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

$3 x^{2} - 0.9 = 0$

Step 4

Find the first derivative.

Differentiate.

By the Sum Rule, the derivative of $x^{3} - 0.9 x + 2$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 0.9 x] + \frac{d}{d x} [2]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 0.9 x) + \frac{d}{d x} (2)$

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

$f' (x) = 3 x^{2} + \frac{d}{d x} (- 0.9 x) + \frac{d}{d x} (2)$

Evaluate $\frac{d}{d x} [- 0.9 x]$ .

Since $- 0.9$ is constant with respect to $x$ , the derivative of $- 0.9 x$ with respect to $x$ is $- 0.9 \frac{d}{d x} [x]$ .

$f' (x) = 3 x^{2} - 0.9 \frac{d}{d x} x + \frac{d}{d x} (2)$

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

$f' (x) = 3 x^{2} - 0.9 \cdot 1 + \frac{d}{d x} (2)$

Multiply $- 0.9$ by $1$ .

$f' (x) = 3 x^{2} - 0.9 + \frac{d}{d x} (2)$

Differentiate using the Constant Rule.

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

$f' (x) = 3 x^{2} - 0.9 + 0$

Add $3 x^{2} - 0.9$ and $0$ .

$f' (x) = 3 x^{2} - 0.9$

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} - 0.9$ .

$3 x^{2} - 0.9$

Step 5

Set the first derivative equal to $0$ then solve the equation $3 x^{2} - 0.9 = 0$ .

Set the first derivative equal to $0$ .

$3 x^{2} - 0.9 = 0$

Add $0.9$ to both sides of the equation.

$3 x^{2} = 0.9$

Divide each term in $3 x^{2} = 0.9$ by $3$ and simplify.

Divide each term in $3 x^{2} = 0.9$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{0.9}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{0.9}{3}$

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

$x^{2} = \frac{0.9}{3}$

Simplify the right side.

Divide $0.9$ by $3$ .

$x^{2} = 0.3$

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

$x = \pm \sqrt{0.3}$

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

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

$x = \sqrt{0.3}$

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

$x = - \sqrt{0.3}$

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

$x = \sqrt{0.3}, - \sqrt{0.3}$

Step 6

Find the values where the derivative is undefined.

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = \sqrt{0.3}, - \sqrt{0.3}$

Step 8

Evaluate the second derivative at $x = \sqrt{0.3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (\sqrt{0.3})$

Step 9

Multiply $6$ by $\sqrt{0.3}$ .

$6 \sqrt{0.3}$

Step 10

$x = \sqrt{0.3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \sqrt{0.3}$ is a local minimum

Step 11

Find the y-value when $x = \sqrt{0.3}$ .

Replace the variable $x$ with $\sqrt{0.3}$ in the expression.

$f (\sqrt{0.3}) = {(\sqrt{0.3})}^{3} - 0.9 \sqrt{0.3} + 2$

Simplify the result.

Simplify each term.

Rewrite ${\sqrt{0.3}}^{3}$ as $\sqrt{{0.3}^{3}}$ .

$f (\sqrt{0.3}) = \sqrt{{0.3}^{3}} - 0.9 \sqrt{0.3} + 2$

Raise $0.3$ to the power of $3$ .

$f (\sqrt{0.3}) = \sqrt{0.027} - 0.9 \sqrt{0.3} + 2$

Multiply $- 0.9$ by $\sqrt{0.3}$ .

$f (\sqrt{0.3}) = \sqrt{0.027} - 0.4929503 + 2$

Simplify by adding terms.

Subtract $0.4929503$ from $\sqrt{0.027}$ .

$f (\sqrt{0.3}) = - 0.32863353 + 2$

Add $- 0.32863353$ and $2$ .

$f (\sqrt{0.3}) = 1.67136646$

The final answer is $1.67136646$ .

$y = 1.67136646$

Step 12

Evaluate the second derivative at $x = - \sqrt{0.3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (- \sqrt{0.3})$

Step 13

Multiply $- 1$ by $6$ .

$- 6 \sqrt{0.3}$

Step 14

$x = - \sqrt{0.3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - \sqrt{0.3}$ is a local maximum

Step 15

Find the y-value when $x = - \sqrt{0.3}$ .

Replace the variable $x$ with $- \sqrt{0.3}$ in the expression.

$f (- \sqrt{0.3}) = {(- \sqrt{0.3})}^{3} - 0.9 (- \sqrt{0.3}) + 2$

Simplify the result.

Simplify each term.

Apply the product rule to $- \sqrt{0.3}$ .

$f (- \sqrt{0.3}) = {(- 1)}^{3} {\sqrt{0.3}}^{3} - 0.9 (- \sqrt{0.3}) + 2$

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

$f (- \sqrt{0.3}) = - {\sqrt{0.3}}^{3} - 0.9 (- \sqrt{0.3}) + 2$

Rewrite ${\sqrt{0.3}}^{3}$ as $\sqrt{{0.3}^{3}}$ .

$f (- \sqrt{0.3}) = - \sqrt{{0.3}^{3}} - 0.9 (- \sqrt{0.3}) + 2$

Raise $0.3$ to the power of $3$ .

$f (- \sqrt{0.3}) = - \sqrt{0.027} - 0.9 (- \sqrt{0.3}) + 2$

Multiply $- 0.9 (- \sqrt{0.3})$ .

Multiply $- 1$ by $- 0.9$ .

$f (- \sqrt{0.3}) = - \sqrt{0.027} + 0.9 \sqrt{0.3} + 2$

Multiply $0.9$ by $\sqrt{0.3}$ .

$f (- \sqrt{0.3}) = - \sqrt{0.027} + 0.4929503 + 2$

Simplify by adding terms.

Add $- \sqrt{0.027}$ and $0.4929503$ .

$f (- \sqrt{0.3}) = 0.32863353 + 2$

Add $0.32863353$ and $2$ .

$f (- \sqrt{0.3}) = 2.32863353$

The final answer is $2.32863353$ .

$y = 2.32863353$

Step 16

These are the local extrema for $f (x) = x^{3} - 0.9 x + 2$ .

$(\sqrt{0.3}, 1.67136646)$ is a local minima

$(- \sqrt{0.3}, 2.32863353)$ is a local maxima

Step 17

Find the Maximum/Minimum Value f(x)=x^3-0.9x+2

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