For which real x-values lays the graph with equation y = -x^4 + 18x^2 - 17 under the x-axis? Thank you!

1 Answer
Nov 7, 2017

#]-oo, -sqrt(17)[uu]-1,1[uu]sqrt(17), +oo[#

Explanation:

1)You must find the zeros of the equation

This is a second-degree equation where the "x" is #x^2#

#x^2=(-18+-sqrt(18^2-4*(-1)(-17)))/(2*(-1))#

#x^2=(-18+-sqrt(324-68))/(-2)#

#x^2=(-18+-sqrt(256))/(-2)#

#x^2=(-18+-16)/-2#

#x^2=9+-8#

#x^2=1 or x^2=17#

Therefore the roots are:

#x=+-1 and x=+-sqrt(17)#

Now we need to know in which direction the polynomium goes in the zeros For this we need the first derivative:

#-4x^3 + 36x #

#-sqrt(17) ->131<0-> uarr#

#-1 ->-32<0 ->darr#

#1 ->32<0->uarr#

#sqrt(17) -><-131 -> 0->darr#

Therefore the polynomial grows in -sqrt(17); decresases in -1; groes in 1 and decreases in sqrt(17) (it gives a form of a M). Taking the obtained zeros we can say that its negative in:

#]-oo, -sqrt(17)[uu]-1,1[uu]sqrt(17), +oo[#