Question

12 multiple choice 1 point find the relative extrema of the function, if they exist. f(x) = x⁴ - 2x² - 8 relative maximum at (0, -8); relative minima at (1, -9), (-1, -9) relative maximum at (0, -8); relative minimum at (1, -9) relative minimum at (0, -8); relative maxima at (1, -9), (-1, 7) relative maximum at (1, -9); relative minimum at (-1, -9)

Explanation:

Step1: Find the first derivative

To find the relative extrema, we first find the first derivative of the function \( f(x) = x^4 - 2x^2 - 8 \). Using the power rule, the derivative \( f'(x) \) is:

$$ f'(x) = 4x^3 - 4x $$

Step2: Find critical points

Set \( f'(x) = 0 \) to find the critical points:

$$ 4x^3 - 4x = 0 $$

Factor out \( 4x \):

$$ 4x(x^2 - 1) = 0 $$

Further factor \( x^2 - 1 \) as a difference of squares:

$$ 4x(x - 1)(x + 1) = 0 $$

Setting each factor equal to zero gives the critical points \( x = 0 \), \( x = 1 \), and \( x = -1 \).

Step3: Find the second derivative

To determine if these critical points are relative maxima or minima, we find the second derivative \( f''(x) \). Differentiating \( f'(x) = 4x^3 - 4x \) gives:

$$ f''(x) = 12x^2 - 4 $$

Step4: Test critical points with the second derivative

• For \( x = 0 \):

Substitute \( x = 0 \) into \( f''(x) \):

$$ f''(0) = 12(0)^2 - 4 = -4 $$

Since \( f''(0) < 0 \), the function has a relative maximum at \( x = 0 \). To find the corresponding \( y \)-value, substitute \( x = 0 \) into \( f(x) \):

$$ f(0) = 0^4 - 2(0)^2 - 8 = -8 $$

So, there is a relative maximum at \( (0, -8) \).

• For \( x = 1 \):

Substitute \( x = 1 \) into \( f''(x) \):

$$ f''(1) = 12(1)^2 - 4 = 12 - 4 = 8 $$

Since \( f''(1) > 0 \), the function has a relative minimum at \( x = 1 \). Substitute \( x = 1 \) into \( f(x) \):

$$ f(1) = 1^4 - 2(1)^2 - 8 = 1 - 2 - 8 = -9 $$

So, there is a relative minimum at \( (1, -9) \).

• For \( x = -1 \):

Substitute \( x = -1 \) into \( f''(x) \):

$$ f''(-1) = 12(-1)^2 - 4 = 12 - 4 = 8 $$

Since \( f''(-1) > 0 \), the function has a relative minimum at \( x = -1 \). Substitute \( x = -1 \) into \( f(x) \):

$$ f(-1) = (-1)^4 - 2(-1)^2 - 8 = 1 - 2 - 8 = -9 $$

So, there is a relative minimum at \( (-1, -9) \).

Answer:

Relative maximum at (0, -8); relative minima at (1, -9), (-1, -9) (the first option)