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Given the function f(x)= e^2x(x^2-2) a. does the decreasing arc reach a local or global minimum? b. does f have a global max?Asked by Anonymous
Given the function f(x)= e^2x(x^2-2)
a. does the decreasing arc reach a local or global minimum?
b. does f have a global max?
a. does the decreasing arc reach a local or global minimum?
b. does f have a global max?
Answers
Answered by
Steve
f = e^2x(x^2-2)
f ' = 2e^2x(x-1)(x+2)
f '' = 2e^2x(2x^2+4x-3)
f'=0 at x=-1,2
f''(-2) < 0, so a max at x = -1
f''(1) > 0, so a min at x=2
Since f(x) --> 0 as x --> -oo, the minimum is a global min.
since f(x) --> +oo as x --> +oo, there is no global max
f ' = 2e^2x(x-1)(x+2)
f '' = 2e^2x(2x^2+4x-3)
f'=0 at x=-1,2
f''(-2) < 0, so a max at x = -1
f''(1) > 0, so a min at x=2
Since f(x) --> 0 as x --> -oo, the minimum is a global min.
since f(x) --> +oo as x --> +oo, there is no global max
Answered by
11
yo do your own hw headass
Answered by
Anonymous
Honestly
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