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?

Question

I got this as the derivative; 15 (4-5 x^2+x^4)

And I got 4 relative maxima.

Answer 1

Come on. There's no way to have two relative maxima without either a minimum or an inflection point in between them.

y' = 15(x^4-5x^2+1)
y" = 30x(2x^2-5)

y" =0 when x = ±√(5/2)
So, f is concave
down in (-∞,-√(5/2))
up in (-√(5/2),0)
down in (0,√(5/2))
up in (√(5/2),+∞)

So, you have two minima and 2 maxima

Check it out at

http://www.wolframalpha.com/input/?i=3x^5+%E2%80%93+25x^3+%2B+60x+%2B+1