Question

How to find maxima and minima. Process
1. to find x where dy/dx=o. let the x =a
2. then find d2y/dx2
3.if d2y/dx2 at x=a >0 then we will find minima for x=a
4.if d2y/dx2 at x=a <0 then we will find maxima for x=a
5.if d2y/dx2 at x=a =0 then we will have to proceed to the d3y/dx3
6.if d3y/dx3 at x=a in not eqal 0 , we will not find any maxima or minima.
7.if d3y/dx3 at x=a equals 0 then we will proceed to d4y/dx4. and if d4y/dx4 at x=a >0 then it's minima and if < 0 then maxima.

Please explain the rule 6 above. Why will we not find any maxima or minima in that case? Please explain with figure and example.

Answer 1

Answer on Question #50714 – Math – Differential Calculus | Equations

How to find maxima and minima. Process

1. to find $x$ where $\frac{dy}{dx} = 0$, let the $x = a$

2. then find $\frac{d^2y}{dx^2}$

3. if $\frac{d^2y}{dx^2} > 0$ at $x = a$, then we will find a minimum for $x = a$

4. if $\frac{d^2y}{dx^2} < 0$ at $x = a$, then we will find a maximum for $x = a$

5. if $\frac{d^2y}{dx^2} = 0$ at $x = a$, then we will have to proceed to the $\frac{d^3y}{dx^3}$

6. if $\frac{d^3y}{dx^3} \neq 0$ at $x = a$, we will not find any maxima or minima.

7. if $\frac{d^3y}{dx^3} = 0$ at $x = a$, then we will proceed to $\frac{d^4y}{dx^4}$ and if $\frac{d^4y}{dx^4} > 0$ at $x = a$, then it's minimum and if $\frac{d^4y}{dx^4} < 0$ at $x = a$, then it is maximum.

Please explain the rule 6 above. Why will we not find any maxima or minima in that case? Please explain with figure and example.

Solution.

6. Let $P$ be the Taylor polynomial of $y(x)$ around $x = a$ of order 3.

Then, $P(x) = k(x - a)^3$ (because $y'(a) = y''(a) = 0$)

and so $y(x) = k(x - a)^3 + O(x - a)^4$.

It follows that sufficiently close to $x = a$ and $y(x)$ has the same sign as $k(x - a)^3$ and behaves like $k(x - a)^3$ and strictly increasing ($k > 0$) or decreasing ($k < 0$) and therefore has not maxima or minima at $x = a$.

7. Let $P$ be the Taylor polynomial of $y(x)$ around $x = a$ of order 4.

Then, $P(x) = k(x - a)^4$ (because $y'(a) = y''(a) = y'''(a) = 0$)

and so $y(x) = k(x - a)^4 + O(x - a)^5$

It follows that sufficiently close to $x = a$ $y(x)$ behaves like $k(x - a)^4$ and has maxima $(k < 0)$ or minima $(k > 0)$ .

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