Calculus Examples

$f (x) = x^{4} - 128 x^{2} + 4096$

Step 1

Find the first derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $x^{4} - 128 x^{2} + 4096$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 128 x^{2}] + \frac{d}{d x} [4096]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 128 x^{2}] + \frac{d}{d x} [4096]$

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

$4 x^{3} + \frac{d}{d x} [- 128 x^{2}] + \frac{d}{d x} [4096]$

Evaluate $\frac{d}{d x} [- 128 x^{2}]$ .

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Since $- 128$ is constant with respect to $x$ , the derivative of $- 128 x^{2}$ with respect to $x$ is $- 128 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} - 128 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [4096]$

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

$4 x^{3} - 128 (2 x) + \frac{d}{d x} [4096]$

Multiply $2$ by $- 128$ .

$4 x^{3} - 256 x + \frac{d}{d x} [4096]$

Differentiate using the Constant Rule.

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Since $4096$ is constant with respect to $x$ , the derivative of $4096$ with respect to $x$ is $0$ .

$4 x^{3} - 256 x + 0$

Add $4 x^{3} - 256 x$ and $0$ .

$4 x^{3} - 256 x$

Step 2

Find the second derivative of the function.

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By the Sum Rule, the derivative of $4 x^{3} - 256 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 256 x]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 256 x)$

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 256 x)$

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

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- 256 x)$

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 256 x)$

Evaluate $\frac{d}{d x} [- 256 x]$ .

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Since $- 256$ is constant with respect to $x$ , the derivative of $- 256 x$ with respect to $x$ is $- 256 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x^{2} - 256 \frac{d}{d x} x$

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

$f'' (x) = 12 x^{2} - 256 \cdot 1$

Multiply $- 256$ by $1$ .

$f'' (x) = 12 x^{2} - 256$

Step 3

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

$4 x^{3} - 256 x = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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Differentiate.

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By the Sum Rule, the derivative of $x^{4} - 128 x^{2} + 4096$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 128 x^{2}] + \frac{d}{d x} [4096]$ .

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 128 x^{2}) + \frac{d}{d x} (4096)$

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

$f' (x) = 4 x^{3} + \frac{d}{d x} (- 128 x^{2}) + \frac{d}{d x} (4096)$

Evaluate $\frac{d}{d x} [- 128 x^{2}]$ .

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Since $- 128$ is constant with respect to $x$ , the derivative of $- 128 x^{2}$ with respect to $x$ is $- 128 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 4 x^{3} - 128 \frac{d}{d x} x^{2} + \frac{d}{d x} (4096)$

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

$f' (x) = 4 x^{3} - 128 (2 x) + \frac{d}{d x} (4096)$

Multiply $2$ by $- 128$ .

$f' (x) = 4 x^{3} - 256 x + \frac{d}{d x} (4096)$

Differentiate using the Constant Rule.

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Since $4096$ is constant with respect to $x$ , the derivative of $4096$ with respect to $x$ is $0$ .

$f' (x) = 4 x^{3} - 256 x + 0$

Add $4 x^{3} - 256 x$ and $0$ .

$f' (x) = 4 x^{3} - 256 x$

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 256 x$ .

$4 x^{3} - 256 x$

Step 5

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - 256 x = 0$ .

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Set the first derivative equal to $0$ .

$4 x^{3} - 256 x = 0$

Factor the left side of the equation.

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Factor $4 x$ out of $4 x^{3} - 256 x$ .

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Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 256 x = 0$

Factor $4 x$ out of $- 256 x$ .

$4 x (x^{2}) + 4 x (- 64) = 0$

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 64)$ .

$4 x (x^{2} - 64) = 0$

Rewrite $64$ as $8^{2}$ .

$4 x (x^{2} - 8^{2}) = 0$

Factor.

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Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 8$ .

$4 x ((x + 8) (x - 8)) = 0$

Remove unnecessary parentheses.

$4 x (x + 8) (x - 8) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 8 = 0$

$x - 8 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 8$ equal to $0$ and solve for $x$ .

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Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Set $x - 8$ equal to $0$ and solve for $x$ .

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Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

The final solution is all the values that make $4 x (x + 8) (x - 8) = 0$ true.

$x = 0, - 8, 8$

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.

$x = 0, - 8, 8$

Step 8

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.

$12 {(0)}^{2} - 256$

Step 9

Evaluate the second derivative.

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Simplify each term.

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

$12 \cdot 0 - 256$

Multiply $12$ by $0$ .

$0 - 256$

Subtract $256$ from $0$ .

$- 256$

Step 10

$x = 0$ 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

Step 11

Find the y-value when $x = 0$ .

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Replace the variable $x$ with $0$ in the expression.

$f (0) = {(0)}^{4} - 128 {(0)}^{2} + 4096$

Simplify the result.

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Simplify each term.

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

$f (0) = 0 - 128 {(0)}^{2} + 4096$

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

$f (0) = 0 - 128 \cdot 0 + 4096$

Multiply $- 128$ by $0$ .

$f (0) = 0 + 0 + 4096$

Simplify by adding numbers.

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Add $0$ and $0$ .

$f (0) = 0 + 4096$

Add $0$ and $4096$ .

$f (0) = 4096$

The final answer is $4096$ .

$y = 4096$

Step 12

Evaluate the second derivative at $x = - 8$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(- 8)}^{2} - 256$

Step 13

Evaluate the second derivative.

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Simplify each term.

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Raise $- 8$ to the power of $2$ .

$12 \cdot 64 - 256$

Multiply $12$ by $64$ .

$768 - 256$

Subtract $256$ from $768$ .

$512$

Step 14

$x = - 8$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 8$ is a local minimum

Step 15

Find the y-value when $x = - 8$ .

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Replace the variable $x$ with $- 8$ in the expression.

$f (- 8) = {(- 8)}^{4} - 128 {(- 8)}^{2} + 4096$

Simplify the result.

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Simplify each term.

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Raise $- 8$ to the power of $4$ .

$f (- 8) = 4096 - 128 {(- 8)}^{2} + 4096$

Raise $- 8$ to the power of $2$ .

$f (- 8) = 4096 - 128 \cdot 64 + 4096$

Multiply $- 128$ by $64$ .

$f (- 8) = 4096 - 8192 + 4096$

Simplify by adding and subtracting.

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Subtract $8192$ from $4096$ .

$f (- 8) = - 4096 + 4096$

Add $- 4096$ and $4096$ .

$f (- 8) = 0$

The final answer is $0$ .

$y = 0$

Step 16

Evaluate the second derivative at $x = 8$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(8)}^{2} - 256$

Step 17

Evaluate the second derivative.

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Simplify each term.

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Raise $8$ to the power of $2$ .

$12 \cdot 64 - 256$

Multiply $12$ by $64$ .

$768 - 256$

Subtract $256$ from $768$ .

$512$

Step 18

$x = 8$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 8$ is a local minimum

Step 19

Find the y-value when $x = 8$ .

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Replace the variable $x$ with $8$ in the expression.

$f (8) = {(8)}^{4} - 128 {(8)}^{2} + 4096$

Simplify the result.

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Simplify each term.

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Raise $8$ to the power of $4$ .

$f (8) = 4096 - 128 {(8)}^{2} + 4096$

Raise $8$ to the power of $2$ .

$f (8) = 4096 - 128 \cdot 64 + 4096$

Multiply $- 128$ by $64$ .

$f (8) = 4096 - 8192 + 4096$

Simplify by adding and subtracting.

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Subtract $8192$ from $4096$ .

$f (8) = - 4096 + 4096$

Add $- 4096$ and $4096$ .

$f (8) = 0$

The final answer is $0$ .

$y = 0$

Step 20

These are the local extrema for $f (x) = x^{4} - 128 x^{2} + 4096$ .

$(0, 4096)$ is a local maxima

$(- 8, 0)$ is a local minima

$(8, 0)$ is a local minima

Step 21