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^4-3x^2?

1 Answer
Feb 26, 2018

x-coordinates of inflection points: x=+-1/sqrt(2)

No discontinuities, all polynomials are continuous.

Explanation:

This function has no discontinuities . A polynomial is always continuous.

To find intervals of concavity and determine inflection points, take the second derivative, set it equal to 0, and solve for x :

y'=4x^3-(2)(3)x=4x^3-6x

y''=(3)(4)x^2-6=12x^2-6

12x^2-6=0

12x^2=6

x^2=6/12=1/2

x=+-sqrt(1/2)=+-1/sqrt(2)

Now, we want to break up the domain of y around these x-values. The domain of y is (-∞,∞). Breaking it up yields:

(-∞,-1/sqrt(2)),(-1/sqrt(2),1/sqrt(2)),(1/sqrt(2),∞)

Now, we want to determine whether y'' is positive or negative in each of these intervals. If y''>0 in an interval, then y is concave up on that interval; if y''<0 on an interval, then y is concave down on that interval. If y'' changes signs at a certain value of x, then there is an inflection point at that x-value.

(-∞,-1/sqrt(2)):

y''(-1)=12(-1)^2-6=12-6>0

y is concave up on (-∞,-1/sqrt(2))

(-1/sqrt(2),1/sqrt(2)):

y''(0)=12(0^2)-6=-6<0

y is concave down on (-1/sqrt(2),1/sqrt(2)) and has changed signs at x=-1/sqrt(2); thus, there is an inflection point at x=-1/sqrt(2).

(1/sqrt(2),∞):

y''(1)=12(1^2)-6=12-6>0

y is concave up on (1/sqrt(2),∞) and has changed signs at x=1/sqrt(2); thus, there is an inflection point at x=1/sqrt(2).