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

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

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Answer 1 · 2017-11-07T17:21:37

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

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:

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

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

1 Answer

Explanation:

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