141.30 how many points of inflection will the graph of the function have

[karush]

If the derivative of a function f is given by
$$f'(x)=\frac{1}{5}(x^2-4)^5-x^2$$
how many points of inflection will the graph of the function have?solution find $f"(x)$
$$f''(x)=2x((x^2-4)^4-1)$$
at $f''(x)=0$ we have factored
$$2 x (x^2 - 5) (x^2 - 3) (x^4 - 8 x^2 + 17) = 0$$
then
$$x=0\quad x=\pm\sqrt{3}\quad \pm\sqrt{5}$$
so we have 5 points of inflectionok I was wondering if this could be solved strictly by observation
also I used the $W\vert A$ to get $ f"(x)$

the only thing I know about finding inflexions is they are zero points of the second direvative of a function
where concave <---> convex

 

[MarkFL]

A point of inflection will always be where the second derivative is zero, but not all places where the second derivative is zero will be a point of inflection. Any roots of the second derivative of even multiplicity will indicate that the second derivative does not change sign across that critical value. For the second derivative, a change in sign implies a change in the direction of concavity (up or down), and that's what is required.

For example, consider the function:

$$f(x)=x^4$$

We find:

$$f''(x)=12x^2$$

This has a root at \(x=0\), but it is of multiplicity 2, and so we know the sign of \(f''\) will not change across this critical value, therefore, the point \((0,0)\) is not a point of inflection.

 

[karush]

thanks I didn't know that

however I assume that a point of inflection may exist even if it is a hole.

 

[MarkFL]

thanks I didn't know that

however I assume that a point of inflection may exist even if it is a hole.

Yes, consider:

$$f(x)=\frac{x^4}{x}$$

The origin is still a point of inflection, even if the function is not defined there. :)