How many inflection points?

[grapz]

The double derivative function is given as:

f'' ( x) = ( x - 1 ) ^3 ( x + 2 ) ( x^2 + 3x + 2 ) ( x ^ 64 - 1 )

How many points of inflection does the graph of f have?

How do i go about doing this.

 

[Deleted member 4993]

The double derivative function is given as:

\(\displaystyle f'' ( x) = ( x - 1 ) ^3 ( x + 2 ) ( x^2 + 3x + 2 ) ( x ^ {64} - 1 )\)

How many points of inflection does the graph of f have?

How do i go about doing this.

How are inflection points and values of f"(x) related?

 

[grapz]

Well, if the double derivative of f(x) changes sign than that means there is an inflection point.

That isn't the problem, , since there is like 10000000000000000 critical points, how do i find out wether the sign of f''(x) changes?

There must be a shortcut, but i can't see it.

thanks

 

[galactus]

There's not as many as you may think. The roots of \(\displaystyle x^{64}-1\) are all complex except for -1 and 1.

 

[grapz]

Thanks for the reply.

I found out that this function has no inflection points

 

[galactus]

Exactly. Take a look at the graph.

At \(\displaystyle f''(1)=0, \;\ f''(-1)=0\), but they are not inflection points. f is concave up on \(\displaystyle (-\infty,\infty)\)

If \(\displaystyle f''(x)>0\) on an open interval, then f is concave up on (a,b)

If \(\displaystyle f''(x)<0\) on an open interval, then f is concave down on (a,b)

 

[Deleted member 4993]

The double derivative function is given as:

\(\displaystyle f'' ( x) = ( x - 1 ) ^3 ( x + 2 ) ( x^2 + 3x + 2 ) ( x ^ {64} - 1 )\)

\(\displaystyle f'' ( x) = ( x - 1 )^4( x + 2 )^2(x + 1)^2 ( x ^ {32} + 1)( x ^ {16} + 1)( x ^ {8} + 1)( x ^ {4} + 1)( x ^ {2} + 1)\)

All the real roots are repeating and of even order - thus f"(x) does not change sign and no inflection point exists for the function in question.

How many points of inflection does the graph of f have?

How do i go about doing this.