How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $y = - x^{5} + 2 x^{3} + 4$ $y=-x^5+2x^3+4$ ?

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How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $y = - x^{5} + 2 x^{3} + 4$ $y=-x^5+2x^3+4$ ?

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Answer 1 · 2018-06-24T20:27:02

No discontinuities.
Points of inflection at $x = 0, \pm \sqrt{\frac{3}{5}}$ $x= 0, +-sqrt(3/5)$
CCU on (inf, $- \sqrt{\frac{3}{5}}$ $-sqrt(3/5)$ ) and (0, $\sqrt{\frac{3}{5}}$ $sqrt(3/5)$ )
CCD on ( $- \sqrt{\frac{3}{5}}$ $-sqrt(3/5)$ ,0) and ( $\sqrt{\frac{3}{5}}$ $sqrt(3/5)$ ,inf)

Explanation:

$f (x)$ $f(x)$ is a polynomial so continuous on reals.

By power rule we can find second derivative to be

$f''(x)=-20x^3+12x$

Setting equal to zero we can find Possible Inflection Points

$-20x^3+12x=0$
$x=0, +-sqrt(3/5)$

Testing concavity between points using $x=-1,-1/2,1/2,1$ gives positive concavity between -inf and $-sqrt(3/5)$ negative concavity from $-sqrt(3/5)$ to 0 positive concavity from 0 to $sqrt(3/5)$ and negative concavity from $sqrt(3/5)$ to inf.

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How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $y = - x^{5} + 2 x^{3} + 4$ $y=-x^5+2x^3+4$ ?

1 Answer

Explanation:

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How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for y=-x^5+2x^3+4y=−x5+2x3+4y=-x^5+2x^3+4?

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Explanation:

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How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $y = - x^{5} + 2 x^{3} + 4$ $y=-x^5+2x^3+4$ ?