Algebra Examples

$f (x) = - 0.01 x^{2} + 0.6 x + 5.8$

Step 1

Find the first derivative of the function.

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By the Sum Rule, the derivative of $- 0.01 x^{2} + 0.6 x + 5.8$ with respect to $x$ is $\frac{d}{d x} [- 0.01 x^{2}] + \frac{d}{d x} [0.6 x] + \frac{d}{d x} [5.8]$ .

$\frac{d}{d x} [- 0.01 x^{2}] + \frac{d}{d x} [0.6 x] + \frac{d}{d x} [5.8]$

Evaluate $\frac{d}{d x} [- 0.01 x^{2}]$ .

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Since $- 0.01$ is constant with respect to $x$ , the derivative of $- 0.01 x^{2}$ with respect to $x$ is $- 0.01 \frac{d}{d x} [x^{2}]$ .

$- 0.01 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [0.6 x] + \frac{d}{d x} [5.8]$

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

$- 0.01 (2 x) + \frac{d}{d x} [0.6 x] + \frac{d}{d x} [5.8]$

Multiply $2$ by $- 0.01$ .

$- 0.02 x + \frac{d}{d x} [0.6 x] + \frac{d}{d x} [5.8]$

Evaluate $\frac{d}{d x} [0.6 x]$ .

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Since $0.6$ is constant with respect to $x$ , the derivative of $0.6 x$ with respect to $x$ is $0.6 \frac{d}{d x} [x]$ .

$- 0.02 x + 0.6 \frac{d}{d x} [x] + \frac{d}{d x} [5.8]$

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

$- 0.02 x + 0.6 \cdot 1 + \frac{d}{d x} [5.8]$

Multiply $0.6$ by $1$ .

$- 0.02 x + 0.6 + \frac{d}{d x} [5.8]$

Differentiate using the Constant Rule.

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Since $5.8$ is constant with respect to $x$ , the derivative of $5.8$ with respect to $x$ is $0$ .

$- 0.02 x + 0.6 + 0$

Add $- 0.02 x + 0.6$ and $0$ .

$- 0.02 x + 0.6$

Step 2

Find the second derivative of the function.

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By the Sum Rule, the derivative of $- 0.02 x + 0.6$ with respect to $x$ is $\frac{d}{d x} [- 0.02 x] + \frac{d}{d x} [0.6]$ .

$f'' (x) = \frac{d}{d x} (- 0.02 x) + \frac{d}{d x} (0.6)$

Evaluate $\frac{d}{d x} [- 0.02 x]$ .

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Since $- 0.02$ is constant with respect to $x$ , the derivative of $- 0.02 x$ with respect to $x$ is $- 0.02 \frac{d}{d x} [x]$ .

$f'' (x) = - 0.02 \frac{d}{d x} x + \frac{d}{d x} (0.6)$

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

$f'' (x) = - 0.02 \cdot 1 + \frac{d}{d x} (0.6)$

Multiply $- 0.02$ by $1$ .

$f'' (x) = - 0.02 + \frac{d}{d x} (0.6)$

Differentiate using the Constant Rule.

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Since $0.6$ is constant with respect to $x$ , the derivative of $0.6$ with respect to $x$ is $0$ .

$f'' (x) = - 0.02 + 0$

Add $- 0.02$ and $0$ .

$f'' (x) = - 0.02$

Step 3

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

$- 0.02 x + 0.6 = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $- 0.01 x^{2} + 0.6 x + 5.8$ with respect to $x$ is $\frac{d}{d x} [- 0.01 x^{2}] + \frac{d}{d x} [0.6 x] + \frac{d}{d x} [5.8]$ .

$f' (x) = \frac{d}{d x} (- 0.01 x^{2}) + \frac{d}{d x} (0.6 x) + \frac{d}{d x} (5.8)$

Evaluate $\frac{d}{d x} [- 0.01 x^{2}]$ .

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Since $- 0.01$ is constant with respect to $x$ , the derivative of $- 0.01 x^{2}$ with respect to $x$ is $- 0.01 \frac{d}{d x} [x^{2}]$ .

$f' (x) = - 0.01 \frac{d}{d x} x^{2} + \frac{d}{d x} (0.6 x) + \frac{d}{d x} (5.8)$

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

$f' (x) = - 0.01 (2 x) + \frac{d}{d x} (0.6 x) + \frac{d}{d x} (5.8)$

Multiply $2$ by $- 0.01$ .

$f' (x) = - 0.02 x + \frac{d}{d x} (0.6 x) + \frac{d}{d x} (5.8)$

Evaluate $\frac{d}{d x} [0.6 x]$ .

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Since $0.6$ is constant with respect to $x$ , the derivative of $0.6 x$ with respect to $x$ is $0.6 \frac{d}{d x} [x]$ .

$f' (x) = - 0.02 x + 0.6 \frac{d}{d x} (x) + \frac{d}{d x} (5.8)$

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

$f' (x) = - 0.02 x + 0.6 \cdot 1 + \frac{d}{d x} (5.8)$

Multiply $0.6$ by $1$ .

$f' (x) = - 0.02 x + 0.6 + \frac{d}{d x} (5.8)$

Differentiate using the Constant Rule.

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Since $5.8$ is constant with respect to $x$ , the derivative of $5.8$ with respect to $x$ is $0$ .

$f' (x) = - 0.02 x + 0.6 + 0$

Add $- 0.02 x + 0.6$ and $0$ .

$f' (x) = - 0.02 x + 0.6$

The first derivative of $f (x)$ with respect to $x$ is $- 0.02 x + 0.6$ .

$- 0.02 x + 0.6$

Step 5

Set the first derivative equal to $0$ then solve the equation $- 0.02 x + 0.6 = 0$ .

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Set the first derivative equal to $0$ .

$- 0.02 x + 0.6 = 0$

Subtract $0.6$ from both sides of the equation.

$- 0.02 x = - 0.6$

Divide each term in $- 0.02 x = - 0.6$ by $- 0.02$ and simplify.

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Divide each term in $- 0.02 x = - 0.6$ by $- 0.02$ .

$\frac{- 0.02 x}{- 0.02} = \frac{- 0.6}{- 0.02}$

Simplify the left side.

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Cancel the common factor of $- 0.02$ .

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Cancel the common factor.

$\frac{- 0.02 x}{- 0.02} = \frac{- 0.6}{- 0.02}$

Divide $x$ by $1$ .

$x = \frac{- 0.6}{- 0.02}$

Simplify the right side.

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Divide $- 0.6$ by $- 0.02$ .

$x = 30$

Step 6

Find the values where the derivative is undefined.

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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.

$x = 30$

Step 8

Evaluate the second derivative at $x = 30$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 0.02$

Step 9

$x = 30$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 30$ is a local maximum

Step 10

Find the y-value when $x = 30$ .

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Replace the variable $x$ with $30$ in the expression.

$f (30) = - 0.01 {(30)}^{2} + 0.6 (30) + 5.8$

Simplify the result.

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Simplify each term.

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Raise $30$ to the power of $2$ .

$f (30) = - 0.01 \cdot 900 + 0.6 (30) + 5.8$

Multiply $- 0.01$ by $900$ .

$f (30) = - 9 + 0.6 (30) + 5.8$

Multiply $0.6$ by $30$ .

$f (30) = - 9 + 18 + 5.8$

Simplify by adding numbers.

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Add $- 9$ and $18$ .

$f (30) = 9 + 5.8$

Add $9$ and $5.8$ .

$f (30) = 14.8$

The final answer is $14.8$ .

$y = 14.8$

Step 11

These are the local extrema for $f (x) = - 0.01 x^{2} + 0.6 x + 5.8$ .

$(30, 14.8)$ is a local maxima

Step 12