Checking answer: Find all local max, min, inflection

[SeekerOfDragons]

Just trying to verify my answer is correct to the following:

Find all local maxima, minima, and points of inflection on

y = x^4 - 2x^2 + 2

y' = 4X^3 - 4X
4X(X^2 - 1) = 0
X = -1, 0, 1

(-1, 1) Local Min
(0, 2) Local Max
(1, 1) Local Min

y'' = 12X^2 - 4
12X^2 = 4
X^2 = 4/12
X = -1/sqrt(3), 1/sqrt(3)

(-1/sqrt(3), 13/9) Inflection Point
1/sqrt(3), 13/9) Inflection Point

to verify the local mins and max, used 1st Derivative test to verify.

Function is decreasing (inf, -1) and (0, 1)
Function is increasing (-1, 0) and (1, inf)

 

[wjm11]

Find all local maxima, minima, and points of inflection on

y = x^4 - 2x^2 + 2

y' = 4X^3 - 4X
4X(X^2 - 1) = 0
X = -1, 0, 1

(-1, 1) Local Min
(0, 2) Local Max
(1, 1) Local Min

y'' = 12X^2 - 4
12X^2 = 4
X^2 = 4/12
X = -1/sqrt(3), 1/sqrt(3)

(-1/sqrt(3), 13/9) Inflection Point
1/sqrt(3), 13/9) Inflection Point

to verify the local mins and max, used 1st Derivative test to verify.

Function is decreasing (inf, -1) and (0, 1)
Function is increasing (-1, 0) and (1, inf)

This all looks good.

Additional info: Besides using the 1st Derivative test, you can also use the 2nd Derivative to predict minima and maxima. When y” is positive, the function is concave “upward” in that region, so if there is a y’ =0 point in that region, it will be a local minimum.

Example: consider x = 1, which was a solution for y’ = 0. Plugging 1 into the 2nd derivative, we get y” = 12(1)^2 – 4 = 8, a positive number. Therefore, the function is concave upward, and the point (1,1) is a local minimum. Make sense?

PS Using a graphing calculator to view the function is also a good way to check your work.

 

[SeekerOfDragons]

thanks for the confirmation. I also did the 2nd test to get concave up (inf, -sqrt(3)/3) and (sqrt(3)/3, inf) and concave down (-sqrt(3)/3, sqrt(3)/3)