hw 20 - absolute extrema section 3.6: problem 5\n(1 point)\nfind the extreme values of the function ( f ) on…

hw 20 - absolute extrema section 3.6: problem 5\n(1 point)\nfind the extreme values of the function ( f ) on the interval ( 0.6,4 ). if an extreme value does not exist, enter dne.\n( f(x)=x^{2}+\frac{2}{x} )\nabsolute minimum value:\nabsolute maximum value:\n(round to three decimal places as needed.)

hw 20 - absolute extrema section 3.6: problem 5\n(1 point)\nfind the extreme values of the function ( f ) on the interval ( 0.6,4 ). if an extreme value does not exist, enter dne.\n( f(x)=x^{2}+\frac{2}{x} )\nabsolute minimum value:\nabsolute maximum value:\n(round to three decimal places as needed.)

Answer

Explanation:

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

Using the power rule, if (f(x)=x^{2}+\frac{2}{x}=x^{2}+2x^{- 1}), then (f^{\prime}(x)=2x - 2x^{-2}=2x-\frac{2}{x^{2}}=\frac{2x^{3}-2}{x^{2}}).

Step2: Find the critical points

Set (f^{\prime}(x) = 0), so (\frac{2x^{3}-2}{x^{2}}=0). Since (x^{2}\neq0) (for (x\in[0.6,4])), then (2x^{3}-2 = 0). [ \begin{align*} 2x^{3}-2&=0\ x^{3}&=1\ x& = 1 \end{align*} ]

Step3: Evaluate the function at the critical point and endpoints

  • Evaluate (f(x)) at (x = 0.6): (f(0.6)=(0.6)^{2}+\frac{2}{0.6}=0.36+\frac{10}{3}\approx0.36 + 3.333=3.693).
  • Evaluate (f(x)) at (x = 1): (f(1)=1^{2}+\frac{2}{1}=1 + 2=3).
  • Evaluate (f(x)) at (x = 4): (f(4)=4^{2}+\frac{2}{4}=16+0.5=16.5).

Answer:

Absolute minimum value: (3.000) Absolute maximum value: (16.500)