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

$] - \infty, - \sqrt{17} [\cup] - 1, 1 [\cup] \sqrt{17}, + \infty [$ $]-oo, -sqrt(17)[uu]-1,1[uu]sqrt(17), +oo[$

Explanation:

1)You must find the zeros of the equation

This is a second-degree equation where the "x" is $x^{2}$ $x^2$

$x^{2} = \frac{- 18 \pm \sqrt{18^{2} - 4 \cdot (- 1) (- 17)}}{2 \cdot (- 1)}$ $x^2=(-18+-sqrt(18^2-4*(-1)(-17)))/(2*(-1))$

$x^{2} = \frac{- 18 \pm \sqrt{324 - 68}}{- 2}$ $x^2=(-18+-sqrt(324-68))/(-2)$

$x^{2} = \frac{- 18 \pm \sqrt{256}}{- 2}$ $x^2=(-18+-sqrt(256))/(-2)$

$x^{2} = \frac{- 18 \pm 16}{- 2}$ $x^2=(-18+-16)/-2$

$x^{2} = 9 \pm 8$ $x^2=9+-8$

$x^{2} = 1 or x^{2} = 17$ $x^2=1 or x^2=17$

Therefore the roots are:

$x = \pm 1 and x = \pm \sqrt{17}$ $x=+-1 and x=+-sqrt(17)$

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

$- 4 x^{3} + 36 x$ $-4x^3 + 36x$

$- \sqrt{17} \to 131 < 0 \to ↑ ⏐$ $-sqrt(17) ->131<0-> uarr$

$- 1 \to - 32 < 0 \to ↓$ $-1 ->-32<0 ->darr$

$1 \to 32 < 0 \to ↑$ $1 ->32<0->uarr$

$\sqrt{17} \to < - 131 \to 0 \to ⏐ ↓$ $sqrt(17) -><-131 -> 0->darr$

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:

$] - \infty, - \sqrt{17} [\cup] - 1, 1 [\cup] \sqrt{17}, + \infty [$ $]-oo, -sqrt(17)[uu]-1,1[uu]sqrt(17), +oo[$

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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

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