Find the stationary points.
<em>f(x)</em> = <em>x</em> + 16/<em>x</em>
<em>f '(x)</em> = 1 - 16/<em>x</em> ²
Solve <em>f</em> <em>'(x)</em> = 0.
1 - 16/<em>x</em> ² = 0 → <em>x</em> ² = 16 → (<em>x</em> - 4) (<em>x</em> + 4) = 0
→ <em>x</em> = -4, <em>x</em> = 4
Check the value of <em>f</em> at the stationary points. <em>x</em> = -4 is not in the provided domain so we can omit it.
<em>f</em> (4) = 8
Check the value of <em>f</em> at the boundary of the domain.
<em>f</em> (0.2) = 401/5 = 80.2
<em>f</em> (16) = 17
Then over [0.2, 16], we have max(<em>f</em> ) = 401/5 and min(<em>f</em> ) = 8.
Answer:
42 trips
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what
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