Question

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

Answer 1

`]-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[`