Calculus Examples

$f (x) = - 5 x^{2} + 14 x + 3$

Find the first derivative of the function.

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By the Sum Rule, the derivative of $- 5 x^{2} + 14 x + 3$ with respect to $x$ is $\frac{d}{d x} [- 5 x^{2}] + \frac{d}{d x} [14 x] + \frac{d}{d x} [3]$ .

$\frac{d}{d x} [- 5 x^{2}] + \frac{d}{d x} [14 x] + \frac{d}{d x} [3]$

Evaluate $\frac{d}{d x} [- 5 x^{2}]$ .

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Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{2}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{2}]$ .

$- 5 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [14 x] + \frac{d}{d x} [3]$

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

$- 5 (2 x) + \frac{d}{d x} [14 x] + \frac{d}{d x} [3]$

Multiply $2$ by $- 5$ .

$- 10 x + \frac{d}{d x} [14 x] + \frac{d}{d x} [3]$

Evaluate $\frac{d}{d x} [14 x]$ .

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Since $14$ is constant with respect to $x$ , the derivative of $14 x$ with respect to $x$ is $14 \frac{d}{d x} [x]$ .

$- 10 x + 14 \frac{d}{d x} [x] + \frac{d}{d x} [3]$

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

$- 10 x + 14 \cdot 1 + \frac{d}{d x} [3]$

Multiply $14$ by $1$ .

$- 10 x + 14 + \frac{d}{d x} [3]$

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$- 10 x + 14 + 0$

Add $- 10 x + 14$ and $0$ .

$- 10 x + 14$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $- 10 x + 14$ with respect to $x$ is $\frac{d}{d x} [- 10 x] + \frac{d}{d x} [14]$ .

$f'' (x) = \frac{d}{d x} (- 10 x) + \frac{d}{d x} (14)$

Evaluate $\frac{d}{d x} [- 10 x]$ .

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Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x$ with respect to $x$ is $- 10 \frac{d}{d x} [x]$ .

$f'' (x) = - 10 \frac{d}{d x} x + \frac{d}{d x} (14)$

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

$f'' (x) = - 10 \cdot 1 + \frac{d}{d x} (14)$

Multiply $- 10$ by $1$ .

$f'' (x) = - 10 + \frac{d}{d x} (14)$

Since $14$ is constant with respect to $x$ , the derivative of $14$ with respect to $x$ is $0$ .

$f'' (x) = - 10 + 0$

Add $- 10$ and $0$ .

$f'' (x) = - 10$

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

$- 10 x + 14 = 0$

Subtract $14$ from both sides of the equation.

$- 10 x = - 14$

Divide each term by $- 10$ and simplify.

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Divide each term in $- 10 x = - 14$ by $- 10$ .

$- \frac{10 x}{- 10} = - \frac{14}{- 10}$

Simplify the left side of the equation by cancelling the common factors.

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Reduce the expression by cancelling the common factors.

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Factor $10$ out of $10 x$ .

$- \frac{10 (x)}{- 10} = - \frac{14}{- 10}$

Move the negative one from the denominator of $\frac{x}{- 1}$ .

$- (- 1 \cdot x) = - \frac{14}{- 10}$

Simplify the expression.

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Multiply $- 1$ by $x$ .

$- (- 1 x) = - \frac{14}{- 10}$

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

$x = - \frac{14}{- 10}$

Simplify the right side of the equation.

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Reduce the expression by cancelling the common factors.

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Factor $2$ out of $14$ .

$x = - \frac{2 (7)}{- 10}$

Cancel the common factors.

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Factor $2$ out of $- 10$ .

$x = - \frac{2 \cdot 7}{2 \cdot - 5}$

Cancel the common factor.

$x = - \frac{2 \cdot 7}{2 \cdot - 5}$

Rewrite the expression.

$x = - \frac{7}{- 5}$

Move the negative in front of the fraction.

$x = \frac{7}{5}$

Evaluate the second derivative at $x = \frac{7}{5}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 10$

$x = \frac{7}{5}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{7}{5}$ is a local maximum

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