Question

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

Answer 1

1) First derivative

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.