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#?

Answer 1

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

No discontinuities, all polynomials are continuous.

A polynomial is always continuous, so this function has no discontinuities.

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).#
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Answer 2
To find the x-coordinates of all points of inflection: 1. Find the second derivative of the function. 2. Set the second derivative equal to zero and solve for x. 3. The solutions obtained will give the x-coordinates of the points of inflection. To find all discontinuities: 1. Examine the function for any potential points of discontinuity, which occur where the function is not defined or exhibits a jump, hole, or vertical asymptote. 2. For the given function, \(y = x^4 - 3x^2\), there are no discontinuities because it is a polynomial function, and polynomial functions are continuous everywhere. To find the open intervals of concavity: 1. Find the second derivative of the function. 2. Determine where the second derivative is positive and negative. 3. The intervals where the second derivative is positive indicate where the function is concave upward, and where it is negative indicates where it is concave downward. For \(y = x^4 - 3x^2\): 1. Find the second derivative: \(y'' = 12x^2 - 6\). 2. Set \(12x^2 - 6 = 0\) and solve for x to find points of inflection. 3. \(12x^2 - 6 = 0 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{\sqrt{2}}{2}\). 4. The points of inflection are at \(x = \frac{\sqrt{2}}{2}\) and \(x = -\frac{\sqrt{2}}{2}\). 5. To determine concavity, test intervals: \((- \infty, -\frac{\sqrt{2}}{2})\), \((-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\), and \((\frac{\sqrt{2}}{2}, \infty)\). 6. Evaluate the second derivative in each interval: - For \(x < -\frac{\sqrt{2}}{2}\), \(y'' > 0\), so the function is concave upward. - For \(-\frac{\sqrt{2}}{2} < x < \frac{\sqrt{2}}{2}\), \(y'' < 0\), so the function is concave downward. - For \(x > \frac{\sqrt{2}}{2}\), \(y'' > 0\), so the function is concave upward. 7. Thus, the open intervals of concavity are \((- \infty, -\frac{\sqrt{2}}{2})\) and \((\frac{\sqrt{2}}{2}, \infty)\).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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