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anonymous

  • 5 years ago

Let f (x) = 4x^(10/3). Find all extreme values (if any) of f on the interval [-8, 8]. Determine at which numbers in the interval these values occur.

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  1. anonymous
    • 5 years ago
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    First Derive f(x). \[f \prime{(x)}=40x^{(7/3)}\] Now find where the derived of f is zero... (extreme point(s)). You do this by setting the derived equal zero. The extreme must be horisontal (0). Solve(0= 40x^{(7/3),x) The global MINIMUM extreme is lokatedf at (0,0) To check it you can enter it on your calculator and have it draw the graph. Notice how the "legs" point up and the extreme is at (0,0). The final answer would be to write it like this: \[f>0 \in ]-\infty;0[\] \[f=0 \in [0]\] \[f<0 \in ]0;\infty[\]

  2. anonymous
    • 5 years ago
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    Thanks Santa's_little_helper; I just solved it a while ago. =D Turns out I just copied some numbers incorrectly. xD Thanks though!

  3. anonymous
    • 5 years ago
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    I forgot about the interval... remember it's not \[-\infty\] but -8 to 8

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