Calculus Examples

$f (x) = 2 x^{4} - 2 x^{3}$

Find the first derivative of the function.

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By the Sum Rule, the derivative of $2 x^{4} - 2 x^{3}$ with respect to $x$ is $\frac{d}{d x} [2 x^{4}] + \frac{d}{d x} [- 2 x^{3}]$ .

$\frac{d}{d x} [2 x^{4}] + \frac{d}{d x} [- 2 x^{3}]$

Evaluate $\frac{d}{d x} [2 x^{4}]$ .

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Since $2$ is constant with respect to $x$ , the derivative of $2 x^{4}$ with respect to $x$ is $2 \frac{d}{d x} [x^{4}]$ .

$2 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 2 x^{3}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$2 (4 x^{3}) + \frac{d}{d x} [- 2 x^{3}]$

Multiply $4$ by $2$ .

$8 x^{3} + \frac{d}{d x} [- 2 x^{3}]$

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

$8 x^{3} - 2 \frac{d}{d x} [x^{3}]$

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

$8 x^{3} - 2 (3 x^{2})$

Multiply $3$ by $- 2$ .

$8 x^{3} - 6 x^{2}$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $8 x^{3} - 6 x^{2}$ with respect to $x$ is $\frac{d}{d x} [8 x^{3}] + \frac{d}{d x} [- 6 x^{2}]$ .

$f'' (x) = \frac{d}{d x} (8 x^{3}) + \frac{d}{d x} (- 6 x^{2})$

Evaluate $\frac{d}{d x} [8 x^{3}]$ .

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Since $8$ is constant with respect to $x$ , the derivative of $8 x^{3}$ with respect to $x$ is $8 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 8 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 6 x^{2})$

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

$f'' (x) = 8 (3 x^{2}) + \frac{d}{d x} (- 6 x^{2})$

Multiply $3$ by $8$ .

$f'' (x) = 24 x^{2} + \frac{d}{d x} (- 6 x^{2})$

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

$f'' (x) = 24 x^{2} - 6 \frac{d}{d x} x^{2}$

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

$f'' (x) = 24 x^{2} - 6 (2 x)$

Multiply $2$ by $- 6$ .

$f'' (x) = 24 x^{2} - 12 x$

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

$8 x^{3} - 6 x^{2} = 0$

Factor $2 x^{2}$ out of $8 x^{3} - 6 x^{2}$ .

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Factor $2 x^{2}$ out of $8 x^{3}$ .

$2 x^{2} (4 x) - 6 x^{2} = 0$

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

$2 x^{2} (4 x) + 2 x^{2} (- 3) = 0$

Factor $2 x^{2}$ out of $2 x^{2} (4 x) + 2 x^{2} (- 3)$ .

$2 x^{2} (4 x - 3) = 0$

Set $2 x^{2}$ equal to $0$ and solve for $x$ .

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Divide each term by $2$ and simplify.

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Divide each term in $2 x^{2} = 0$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{0}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{0}{2}$

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

$x^{2} = \frac{0}{2}$

Divide $0$ by $2$ .

$x^{2} = 0$

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

$x = \pm \sqrt{0}$

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

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Simplify the right side of the equation.

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Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

$\pm 0$ is equal to $0$ .

$x = 0$

Set $4 x - 3$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$4 x - 3 = 0$

Add $3$ to both sides of the equation.

$4 x = 3$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = 3$ by $4$ .

$\frac{4 x}{4} = \frac{3}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 x}{4} = \frac{3}{4}$

Divide $x$ by $1$ .

$x = \frac{3}{4}$

The solution is the result of $2 x^{2} = 0$ and $4 x - 3 = 0$ .

$x = 0, \frac{3}{4}$

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.

$24 {(0)}^{2} - 12 \cdot 0$

Evaluate the second derivative.

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Simplify each term.

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Raising $0$ to any positive power yields $0$ .

$24 \cdot 0 - 12 \cdot 0$

Multiply $24$ by $0$ .

$0 - 12 \cdot 0$

Multiply $- 12$ by $0$ .

$0 + 0$

Add $0$ and $0$ .

$0$

Apply the first derivative test.

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Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, 0) \cup (0, 0.75) \cup (0.75, \infty)$

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $8 x^{3} - 6 x^{2}$ to check if the result is negative or positive.

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Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = 8 {(- 2)}^{3} - 6 {(- 2)}^{2}$

Simplify the result.

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Simplify each term.

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Raise $- 2$ to the power of $3$ .

$f' (- 2) = 8 \cdot - 8 - 6 {(- 2)}^{2}$

Multiply $8$ by $- 8$ .

$f' (- 2) = - 64 - 6 {(- 2)}^{2}$

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

$f' (- 2) = - 64 - 6 \cdot 4$

Multiply $- 6$ by $4$ .

$f' (- 2) = - 64 - 24$

Subtract $24$ from $- 64$ .

$f' (- 2) = - 88$

The final answer is $- 88$ .

$- 88$

Substitute any number, such as $0.38$ , from the interval $(0, 0.75)$ in the first derivative $8 x^{3} - 6 x^{2}$ to check if the result is negative or positive.

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Replace the variable $x$ with $0.38$ in the expression.

$f' (0.38) = 8 {(0.38)}^{3} - 6 {(0.38)}^{2}$

Simplify the result.

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Simplify each term.

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Raise $0.38$ to the power of $3$ .

$f' (0.38) = 8 \cdot 0.054872 - 6 {(0.38)}^{2}$

Multiply $8$ by $0.054872$ .

$f' (0.38) = 0.438976 - 6 {(0.38)}^{2}$

Raise $0.38$ to the power of $2$ .

$f' (0.38) = 0.438976 - 6 \cdot 0.1444$

Multiply $- 6$ by $0.1444$ .

$f' (0.38) = 0.438976 - 0.8664$

Subtract $0.8664$ from $0.438976$ .

$f' (0.38) = - 0.427424$

The final answer is $- 0.427424$ .

$- 0.427424$

Substitute any number, such as $3$ , from the interval $(0.75, \infty)$ in the first derivative $8 x^{3} - 6 x^{2}$ to check if the result is negative or positive.

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Replace the variable $x$ with $3$ in the expression.

$f' (3) = 8 {(3)}^{3} - 6 {(3)}^{2}$

Simplify the result.

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Simplify each term.

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

$f' (3) = 8 \cdot 27 - 6 {(3)}^{2}$

Multiply $8$ by $27$ .

$f' (3) = 216 - 6 {(3)}^{2}$

Raise $3$ to the power of $2$ .

$f' (3) = 216 - 6 \cdot 9$

Multiply $- 6$ by $9$ .

$f' (3) = 216 - 54$

Subtract $54$ from $216$ .

$f' (3) = 162$

The final answer is $162$ .

$162$

Since the first derivative did not change signs around $x = 0$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Since the first derivative changed signs from negative to positive around $x = 0.75$ , then $x = 0.75$ is a local minimum.

$x = 0.75$ is a local minimum

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