How many points of inflection does the function f(x) = x^8 - x^2 have?

Answer 1

f(x) = x^8 - x^2

To solve for the points of inflection, we need to find the second derivative and solve for our critical values.

But we need the first derivative to get to the second, so

f'(x) = 8x^7 - 2x

Differentiating again,

f''(x) = 56x^6 - 2

Making f''(x) = 0 will give us our critical values.

0 = 56x^6 - 2. Dividing by 2,
0 = 28x^6 - 1. Factoring as an (imperfect) different of squares, we get

0 = (√(28)x^3 - 1)(√(28)x^3 + 1)

Equating each to 0, we get

√(28)x^3 - 1 = 0
√(28)x^3 + 1 = 0

√(28)x^3 = 1
√(28)x^3 = -1

x^3 = 1/√(28)
x^3 = -1/√(28)

It follows that

x = 1/ [ √(28) ]^(1/3)
x = -1/ [ √(28) ]^(1/3)

Are these actually inflection points? The only way to know for sure is to make a number line, and test f''(x) = 56x^6 - 2 for
positivity/negativity in each region.

. . . . . . (-1/ [ √(28) ]^(1/3)) . . . . . . (-1/ [ √(28) ]^(1/3)) . . . . . . .

f''(x) = 56x^6 - 2 = 2(28x^6 - 1)

For the first region, test x = -100. Then 2(28x^6 - 1) is definitely a positive number. Mark the region as positive.

. .{+}. . . (-1/ [ √(28) ]^(1/3)) . . . . . . (-1/ [ √(28) ]^(1/3)) . . . . . . .

For the second region, test x = 0. Then 2(28x^6 - 1) is negative. Mark the region as negative.

. .{+}. . . (-1/ [ √(28) ]^(1/3)) . .{-}. . . (-1/ [ √(28) ]^(1/3)) . . . . .

For the third region, test x = 100. We get a positive number as well.

. .{+}. . . (-1/ [ √(28) ]^(1/3)) . .{-}. . . (-1/ [ √(28) ]^(1/3)) . . {+} . .

Conclusion:

f(x) is concave up on the positive intervals; on
(-infinity, (-1/ [ √(28) ]^(1/3)) ) U ( (-1/ [ √(28) ]^(1/3)) , infinity)

f(x) is concave down on
( -1/ [ √(28) ]^(1/3) , 1/ [ √(28) ]^(1/3) )

Two sign changes on the number line implies that f(x) has two inflection points.

Answer 2

You require f '(x) = 0 and f ''(x) = 0 and f '''(x) not = 0

f '(x) = 8x^7 - 2x
= 0 for tp
2x(x^6 - 1) = 0 so x = 0 or x^6 = 1/4

f ''(x) = 56x^6 - 2

when x = 0 f ''(x) = -2 not an inflection
when x^6 = 1/4 f ''(x) = 12 not an inflection

There appear to be no points of inflection on this curve

Points of inflection are those where the gradient actually reaches zero at the point of inflection ...... + 0 + or - 0 -

You have two points of oblique inflection but I don't know if you are considering these or not. These are where the gradient goes + less+ + or - less- - through the inflection.
These will occur at f ''(x) = 0 x^6 = 1/28 x = + or - 6th root of 1/28

Answer 3

Just find the second derivative:

f prime=8x^7-2x

f double prime=56x^6-2

which is the same as 2(28x^6-1)

so basically you have to find the root of 28x^6-1 so you make it equal zero and solve for x (the root or roots are the critical numbers, which you need to eventually find the inflection points.)

28x^6-1=0
28x^6=1
x^6=1/28
x=1/28^(1/6)
x=-1/28^(1/6)

So basically it has two inflection points (at x=+-1/28^(1/6))

Answer 4

f(x) = x^8 - x^2
f'(x) = 8x^7 - 2x
f''(x) = 56x^6 - 2 = 0
x^6 - 1/28 = 0
(x^3 + √(1/28)) (x^3 - √(1/28)) = 0
There are only two real roots here, so there are only two points of inflection.

Answer 5

2. Points of inflection are not extrema. Just imagine the function. For small x it is a parabola opening downwards. For large x, it is parabola-like opening upwards. Thus, it goes from concave up to concave down back to concave up as you go from -infinity to +infinity