Question

find the absolute maximum and the absolute minimum for the function given. give your answers correct to at least 3 decimal places.
$f(x) = x^8 e^{-x^2}$ on $-4, 4$.

absolute minimum:
is
occurs at

absolute maximum:
is
occurs at

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Explanation:

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

To find critical points, we first take the derivative of \( f(x) = x^8 e^{-x^2} \) using the product rule \( (uv)' = u'v + uv' \), where \( u = x^8 \) and \( v = e^{-x^2} \).

• \( u' = 8x^7 \)
• \( v' = e^{-x^2} \cdot (-2x) \) (by chain rule)

So, \( f'(x) = 8x^7 e^{-x^2} + x^8 e^{-x^2}(-2x) = x^7 e^{-x^2}(8 - 2x^2) \).

Step2: Find critical points

Set \( f'(x) = 0 \). Since \( e^{-x^2} > 0 \) for all \( x \), we solve:
\( x^7(8 - 2x^2) = 0 \)
This gives \( x^7 = 0 \) (so \( x = 0 \)) or \( 8 - 2x^2 = 0 \) (so \( x^2 = 4 \), \( x = \pm 2 \)). Also, check endpoints \( x = -4, 4 \).

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

• At \( x = -4 \): \( f(-4) = (-4)^8 e^{-(-4)^2} = 65536 e^{-16} \approx 65536 \cdot 1.1254 \times 10^{-7} \approx 0.00737 \)
• At \( x = -2 \): \( f(-2) = (-2)^8 e^{-(-2)^2} = 256 e^{-4} \approx 256 \cdot 0.01832 \approx 4.690 \)
• At \( x = 0 \): \( f(0) = 0^8 e^{0} = 0 \)
• At \( x = 2 \): \( f(2) = 2^8 e^{-4} = 256 e^{-4} \approx 4.690 \) (same as \( x = -2 \) due to even function)
• At \( x = 4 \): \( f(4) = 4^8 e^{-16} = 65536 e^{-16} \approx 0.00737 \) (same as \( x = -4 \))

Answer:

Absolute minimum: is \( 0.000 \) (exactly 0 at \( x = 0 \)), occurs at \( x = 0.000 \)
Absolute maximum: is \( 4.690 \), occurs at \( x = \pm 2.000 \)

(Note: For \( x = -2 \) and \( x = 2 \), \( f(x) \) is equal. Rounded to 3 decimals, \( e^{-4} \approx 0.0183156 \), so \( 256 \times 0.0183156 \approx 4.690 \). At \( x = 0 \), \( f(x) = 0 \). At \( x = \pm 4 \), \( f(x) \approx 0.007 \), which is less than 4.690 but greater than 0.)

Final answers:
Absolute minimum: is \( \boldsymbol{0.000} \), occurs at \( \boldsymbol{0.000} \)
Absolute maximum: is \( \boldsymbol{4.690} \), occurs at \( \boldsymbol{\pm 2.000} \)