Calculus Examples

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

Find the first derivative of the function.

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

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

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

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

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

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

Multiply $3$ by $- 2$ .

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

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

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

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

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

Multiply $2$ by $- 3$ .

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

Find the second derivative of the function.

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

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

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

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

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

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

Multiply $2$ by $- 6$ .

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

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

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

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

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

$f'' (x) = - 12 x - 6 \cdot 1$

Multiply $- 6$ by $1$ .

$f'' (x) = - 12 x - 6$

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

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

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

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

$- 6 x \cdot x - 6 x = 0$

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

$- 6 x \cdot x - 6 x \cdot 1 = 0$

Factor $- 6 x$ out of $- 6 x (x) - 6 x (1)$ .

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

Divide each term by $- 6$ and simplify.

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Divide each term in $- 6 x = 0$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{0}{- 6}$

Cancel the common factor of $- 6$ .

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Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{0}{- 6}$

Divide $x$ by $1$ .

$x = \frac{0}{- 6}$

Divide $0$ by $- 6$ .

$x = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The solution is the result of $- 6 x = 0$ and $x + 1 = 0$ .

$x = 0, - 1$

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 \cdot 0 - 6$

Evaluate the second derivative.

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Multiply $- 12$ by $0$ .

$0 - 6$

Subtract $6$ from $0$ .

$- 6$

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

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.

$- 12 \cdot - 1 - 6$

Evaluate the second derivative.

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Multiply $- 12$ by $- 1$ .

$12 - 6$

Subtract $6$ from $12$ .

$6$

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

$x = - 1$ is a local minimum

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

$x = 0$ is a local maximum

$x = - 1$ is a local minimum

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