Question

find the value(s) of p for which the function ?

A function `f(x)` is defined about a point `x = 0` by
`f(x) = x + px^2 + O(x^5)`,
where `p` `in` `RR`. Apply Taylor polynomial about `x = 0` to find the value(s) of `p` for which the function
`g(x) = f(sinx) - x +` `(1)/(3)x^2`+ `(1)/(3)x^3`,

(i) Has a local minimum value ?
(ii) Has a local maximum value ?
(iii) Has an inflection point at `x = 0` ?

Answer 1

(i) local minimum if `p > -1/3`
(ii) local maximum if `p < -1/3`
(i) inflection point if `p = -1/3`

Explanation:

Let's first look at the first few terms of the Taylor expansion of the function `f(sin x)` about `x=0`. We can deduce this easily from the given consition :

`f(sin x) = f(x-x^3/(3!)+x^5/(5!)+...)`
`qquad = (x-x^3/(3!)+x^5/(5!)+...)+p(x-x^3/(3!)+x^5/(5!)+...)^2+O(x^5)`
`qquad = (x-x^3/(3!)+x^5/(5!)+...)+p(x^2-2x^4/(3!)+...)+O(x^5)`
`qquad = x+px^2-x^3/(3!)-2px^4/(3!)+O(x^5)`

Let us now take a look at the Taylor expansion for `g(x)`:

`g(x) = f(sin x)-x+x^2/3+x^3/3`
`qquad =( x+px^2-x^3/(3!)-2px^4/(3!)+O(x^5))-x+x^2/3+x^3/3`
`qquad = (p+1/3)x^2+x^3/6-p/12 x^4+O(x^5)`

Thus if `p+1/3` is non-zero, the behavior of `g(x)` very close to `x=0` is dominated by the term

`(p+1/3)x^2`

Since `x^2` has a local minimum at `x=0`, `g(x)` will have a local minimum at the same point if

`p+1/3>0 implies p > -1/3`

On the other hand,

`p < -1/3`

will imply that the function will have a local maximum at `x=0`..

If `p=1/3`, the behavior of `g(x)` very close to `x=0` is dominated by next term `x^3/6`. As this increases from `g(0)=0` when `x` increases from 0, and decreases from `g(0)` as `x` decreases from 0, the point `x=0` will be a point of inflection in this case.

Note

This could also have been solved by using successive derivatives of `g(x)` at `x=0` - but the criteria for maxima/minima/inflexion is based actually on the Taylor series - so this approach is logically better.