Question

consider the function $f(x) = 2x^3 + 6x^2 - 144x + 1$, $-6 \leq x \leq 5$. this function has an absolute minimum value equal to \boxed{} and an absolute maximum value equal to \boxed{}

Explanation:

Step1: Find the derivative of \( f(x) \)

To find critical points, we first take the derivative of \( f(x) = 2x^3 + 6x^2 - 144x + 1 \). Using the power rule, \( f'(x) = 6x^2 + 12x - 144 \).

Step2: Solve \( f'(x) = 0 \) for critical points

Set \( 6x^2 + 12x - 144 = 0 \). Divide both sides by 6: \( x^2 + 2x - 24 = 0 \). Factor the quadratic: \( (x + 6)(x - 4) = 0 \). So critical points are \( x = -6 \) and \( x = 4 \).

Step3: Evaluate \( f(x) \) at critical points and endpoints

The interval is \( -6 \leq x \leq 5 \), so endpoints are \( x = -6 \) and \( x = 5 \), and critical point \( x = 4 \) (since \( x = -6 \) is also an endpoint).

• For \( x = -6 \): \( f(-6) = 2(-6)^3 + 6(-6)^2 - 144(-6) + 1 = 2(-216) + 6(36) + 864 + 1 = -432 + 216 + 864 + 1 = 649 \).
• For \( x = 4 \): \( f(4) = 2(4)^3 + 6(4)^2 - 144(4) + 1 = 2(64) + 6(16) - 576 + 1 = 128 + 96 - 576 + 1 = -351 \).
• For \( x = 5 \): \( f(5) = 2(5)^3 + 6(5)^2 - 144(5) + 1 = 2(125) + 6(25) - 720 + 1 = 250 + 150 - 720 + 1 = -319 \).

Step4: Determine absolute min and max

Compare the values: \( f(-6) = 649 \), \( f(4) = -351 \), \( f(5) = -319 \). The smallest is \( -351 \) (absolute min), the largest is \( 649 \) (absolute max).

Answer:

Absolute minimum value: \(-351\)
Absolute maximum value: \(649\)