Find the Local Maxima and Minima 2x^4-4x^2+1

$2 x^{4} - 4 x^{2} + 1$

Step 1

Write $2 x^{4} - 4 x^{2} + 1$ as a function.

$f (x) = 2 x^{4} - 4 x^{2} + 1$

Step 2

Find the first derivative of the function.

By the Sum Rule, the derivative of $2 x^{4} - 4 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [2 x^{4}] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [2 x^{4}] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [1]$

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

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

$2 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [1]$

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

$2 (4 x^{3}) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [1]$

Multiply $4$ by $2$ .

$8 x^{3} + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [1]$

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

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

$8 x^{3} - 4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

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

$8 x^{3} - 4 (2 x) + \frac{d}{d x} [1]$

Multiply $2$ by $- 4$ .

$8 x^{3} - 8 x + \frac{d}{d x} [1]$

Differentiate using the Constant Rule.

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$8 x^{3} - 8 x + 0$

Add $8 x^{3} - 8 x$ and $0$ .

$8 x^{3} - 8 x$

Step 3

Find the second derivative of the function.

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

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

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

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

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

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

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

Multiply $3$ by $8$ .

$f'' (x) = 24 x^{2} + \frac{d}{d x} (- 8 x)$

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

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

$f'' (x) = 24 x^{2} - 8 \frac{d}{d x} x$

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

$f'' (x) = 24 x^{2} - 8 \cdot 1$

Multiply $- 8$ by $1$ .

$f'' (x) = 24 x^{2} - 8$

Step 4

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

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

Step 5

Find the first derivative.

By the Sum Rule, the derivative of $2 x^{4} - 4 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [2 x^{4}] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [1]$ .

$f' (x) = \frac{d}{d x} (2 x^{4}) + \frac{d}{d x} (- 4 x^{2}) + \frac{d}{d x} (1)$

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

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

$f' (x) = 2 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 4 x^{2}) + \frac{d}{d x} (1)$

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

$f' (x) = 2 (4 x^{3}) + \frac{d}{d x} (- 4 x^{2}) + \frac{d}{d x} (1)$

Multiply $4$ by $2$ .

$f' (x) = 8 x^{3} + \frac{d}{d x} (- 4 x^{2}) + \frac{d}{d x} (1)$

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

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

$f' (x) = 8 x^{3} - 4 \frac{d}{d x} x^{2} + \frac{d}{d x} (1)$

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

$f' (x) = 8 x^{3} - 4 (2 x) + \frac{d}{d x} (1)$

Multiply $2$ by $- 4$ .

$f' (x) = 8 x^{3} - 8 x + \frac{d}{d x} (1)$

Differentiate using the Constant Rule.

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f' (x) = 8 x^{3} - 8 x + 0$

Add $8 x^{3} - 8 x$ and $0$ .

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

The first derivative of $f (x)$ with respect to $x$ is $8 x^{3} - 8 x$ .

$8 x^{3} - 8 x$

Step 6

Set the first derivative equal to $0$ then solve the equation $8 x^{3} - 8 x = 0$ .

Set the first derivative equal to $0$ .

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

Factor the left side of the equation.

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

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

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

Factor $8 x$ out of $- 8 x$ .

$8 x (x^{2}) + 8 x (- 1) = 0$

Factor $8 x$ out of $8 x (x^{2}) + 8 x (- 1)$ .

$8 x (x^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$8 x (x^{2} - 1^{2}) = 0$

Factor.

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 $b = 1$ .

$8 x ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$8 x (x + 1) (x - 1) = 0$

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

$x = 0$

$x + 1 = 0$

$x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $8 x (x + 1) (x - 1) = 0$ true.

$x = 0, - 1, 1$

Step 7

Find the values where the derivative is undefined.

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 8

Critical points to evaluate.

$x = 0, - 1, 1$

Step 9

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.

$24 {(0)}^{2} - 8$

Step 10

Evaluate the second derivative.

Simplify each term.

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

$24 \cdot 0 - 8$

Multiply $24$ by $0$ .

$0 - 8$

Subtract $8$ from $0$ .

$- 8$

Step 11

$x = 0$ 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 12

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

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

$f (0) = 2 {(0)}^{4} - 4 {(0)}^{2} + 1$

Simplify the result.

Simplify each term.

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

$f (0) = 2 \cdot 0 - 4 {(0)}^{2} + 1$

Multiply $2$ by $0$ .

$f (0) = 0 - 4 {(0)}^{2} + 1$

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

$f (0) = 0 - 4 \cdot 0 + 1$

Multiply $- 4$ by $0$ .

$f (0) = 0 + 0 + 1$

Simplify by adding numbers.

Add $0$ and $0$ .

$f (0) = 0 + 1$

Add $0$ and $1$ .

$f (0) = 1$

The final answer is $1$ .

$y = 1$

Step 13

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.

$24 {(- 1)}^{2} - 8$

Step 14

Evaluate the second derivative.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$24 \cdot 1 - 8$

Multiply $24$ by $1$ .

$24 - 8$

Subtract $8$ from $24$ .

$16$

Step 15

$x = - 1$ 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

Step 16

Find the y-value when $x = - 1$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = 2 {(- 1)}^{4} - 4 {(- 1)}^{2} + 1$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $4$ .

$f (- 1) = 2 \cdot 1 - 4 {(- 1)}^{2} + 1$

Multiply $2$ by $1$ .

$f (- 1) = 2 - 4 {(- 1)}^{2} + 1$

Raise $- 1$ to the power of $2$ .

$f (- 1) = 2 - 4 \cdot 1 + 1$

Multiply $- 4$ by $1$ .

$f (- 1) = 2 - 4 + 1$

Simplify by adding and subtracting.

Subtract $4$ from $2$ .

$f (- 1) = - 2 + 1$

Add $- 2$ and $1$ .

$f (- 1) = - 1$

The final answer is $- 1$ .

$y = - 1$

Step 17

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.

$24 {(1)}^{2} - 8$

Step 18

Evaluate the second derivative.

Simplify each term.

One to any power is one.

$24 \cdot 1 - 8$

Multiply $24$ by $1$ .

$24 - 8$

Subtract $8$ from $24$ .

$16$

Step 19

$x = 1$ 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

Step 20

Find the y-value when $x = 1$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = 2 {(1)}^{4} - 4 {(1)}^{2} + 1$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 2 \cdot 1 - 4 {(1)}^{2} + 1$

Multiply $2$ by $1$ .

$f (1) = 2 - 4 {(1)}^{2} + 1$

One to any power is one.

$f (1) = 2 - 4 \cdot 1 + 1$

Multiply $- 4$ by $1$ .

$f (1) = 2 - 4 + 1$

Simplify by adding and subtracting.

Subtract $4$ from $2$ .

$f (1) = - 2 + 1$

Add $- 2$ and $1$ .

$f (1) = - 1$

The final answer is $- 1$ .

$y = - 1$

Step 21

These are the local extrema for $f (x) = 2 x^{4} - 4 x^{2} + 1$ .

$(0, 1)$ is a local maxima

$(- 1, - 1)$ is a local minima

$(1, - 1)$ is a local minima

Step 22

Find the Local Maxima and Minima 2x^4-4x^2+1

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