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witerose701
04.08.2019 •
Mathematics
Consider the function y=3x^5-25x^3+60x+1. use the first or second derivative test to test the critical points. how many relative maxima did you find? a)0 b)1 c)2 d)3 e)4
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Ответ:
dy / dx = 15x^4 - 75x^2 + 60
relative maxima => dy / dx = 0
=> 15x^4 - 75x^2 + 60 = 0
=> x^4 - 5x^2 + 4 = 0
Solve by factoring: (x^2 - 4) (x^2 - 1) = 0
=> x^2 = 4 => x = 2 and x = -2
x^2 = 1 => x = 1 and x = -1
So, there are four candidates -4, -1, 1 and 4.
To conclude whether they are relative maxima you must take the second derivative. Only those whose second derivative is negative are relative maxima.
2) Second derivative:
d^2 y / dx^2 = 60x^3 - 150x
Factor: 30x(2x^2 - 5)
The second derivative test tells that the second derivative in a relative maxima is less than zero.
=> 30x (2x^2 - 5) < 0
=> x < 0 and x^2 > 5/2
=> x < 0 and x < √(5/2) or x > √(5/2)
=> x = - 2 is a local maxima
Also, x > 0 and x^2 < 5/2 is a solution
=> x = 1 is other local maxima.
Those are the only two local maxima.
The answer is the option c) 2 local maxima.
Ответ:
21 per min
Step-by-step explanation: