How do you find points of inflection and determine the intervals of concavity given #y=2x^4-4x^2+1#?

Answer 1

inflection points #(sqrt(3)/3, -1/9), (-sqrt(3)/3, -1/9)#
concave up #(-oo, -sqrt(3)/3), (sqrt(3)/3, oo)#
concave down #(-sqrt(3)/3, sqrt(3)/3)#

Given: #y = 2x^4 -4x^2 + 1#
Infection points can be found by setting #y'' = 0#; but first you need to know if you have minimums and/or maximums. You can use the first derivative test or the second derivative test. The second derivative test is easier if the second derivative is easy to find. However, it has a restriction. If #y''(critical" "value)" = 0# you will then need to use the first derivative test.

Using the second derivative test, let's:

Find the first derivative: #y' = 8x^3 - 8x = 0#
Factor: #8x(x^2-1) = 8x(x - 1)(x + 1) = 0#
critical values: #x = 0, x = +-1#
Find the second derivative: #y'' = 24x^2 -8#

Determine the relative highest and lowest points:

# y''(0) < 0# relative maximum at #x = 0#
#y''(-1) > 0# relative minimum at #x = -1#
#y''(1) > 0# relative minimum at #x = 1#

Locate the inflection points:

#y'' = 24x^2 -8 = 0#
Factor: #8(3x^2 - 1) = 0#
#3x^2 = 1#
#x^2 = 1/3#
#x = +- sqrt(1/3) = +- 1/(sqrt(3)) = +- (sqrt(3))/3#
#y((sqrt(3))/3) = 2((sqrt(3))/3)^4 - 4((sqrt(3))/3)^2 +1 = -1/9#
#y(-(sqrt(3))/3) = 2((sqrt(3))/3)^4 - 4((sqrt(3))/3)^2 +1 = -1/9#
inflection points #(sqrt(3)/3, -1/9), (-sqrt(3)/3, -1/9)#
Concave down is when #y'# is decreasing. A relative maximum would be concave down to the point of inflection.
Concave up is when #y'# is increasing. A relative minimum would be concave up to the point of inflection.

Concavity intervals:

#(-oo, -sqrt(3)/3), (-sqrt(3)/3. sqrt(3)/3), (sqrt(3)/3, oo)#
remember from work above, that there is a relative maximum at #x = 0# and two relative minimums at #x = +-1#
concave up #(-oo, -sqrt(3)/3), (sqrt(3)/3, oo)# concave down #(-sqrt(3)/3, sqrt(3)/3)#
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Answer 2

To find points of inflection and determine the intervals of concavity for the function ( y = 2x^4 - 4x^2 + 1 ), follow these steps:

  1. Find the second derivative of the function.
  2. Set the second derivative equal to zero and solve for ( x ) to find the potential points of inflection.
  3. Test the intervals between these points of inflection using the second derivative test to determine the concavity of the function.

Let's go through each step:

  1. Find the second derivative of the function ( y = 2x^4 - 4x^2 + 1 ): [ y' = 8x^3 - 8x ] [ y'' = 24x^2 - 8 ]

  2. Set the second derivative equal to zero and solve for ( x ) to find potential points of inflection: [ 24x^2 - 8 = 0 ] [ 24x^2 = 8 ] [ x^2 = \frac{8}{24} = \frac{1}{3} ] [ x = \pm \sqrt{\frac{1}{3}} ]

  3. Test the intervals between these points of inflection:

    • Choose a value of ( x ) in each interval and plug it into the second derivative.
    • If ( y'' > 0 ), the function is concave up in that interval.
    • If ( y'' < 0 ), the function is concave down in that interval.

Choose ( x = 0 ) for the interval ( (-\infty, -\sqrt{\frac{1}{3}}) ): [ y'' = 24(0)^2 - 8 = -8 < 0 ] So, the function is concave down in this interval.

Choose ( x = \frac{1}{2} ) for the interval ( (-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}) ): [ y'' = 24\left(\frac{1}{2}\right)^2 - 8 = 6 - 8 = -2 < 0 ] So, the function is concave down in this interval.

Choose ( x = 1 ) for the interval ( (\sqrt{\frac{1}{3}}, \infty) ): [ y'' = 24(1)^2 - 8 = 16 > 0 ] So, the function is concave up in this interval.

Therefore, the points of inflection are ( (-\sqrt{\frac{1}{3}}, \frac{1}{2}) ) and ( (\sqrt{\frac{1}{3}}, \frac{1}{2}) ), and the intervals of concavity are ( (-\infty, -\sqrt{\frac{1}{3}}) ) and ( (\sqrt{\frac{1}{3}}, \infty) ), where the function is concave down, and ( (-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}) ), where the function is concave up.

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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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