how to find the minimum points of a equation?

how to find the minimum points of y=x^4-8^2+3 ?
do i need to differentiate it first?

many thanks!

You do need to differentiate the equation as this gives us a gradient function. So at a minimum point the gradient is 0. So you differentiate the equation and set it equal to 0. You then solve by factorising the cubic equation. And solvin for x. To see if it is a maximum point or minimum point (or inflection point) you then take the second derivative of the equation - by differentiating the derivative. Then plug in the x values you got for dy/dx = 0 and if the second derivative is greater than 0 then it is a min point at that x value. If the second derivative is less than 0 then it is a max point at that x value. If the second derivative is 0 then you have a point of inflection.

You need to use the fact that $\frac{dy}{dx}=0$ at minima.

how to find the minimum points of y=x^4-8^2+3 ?
do i need to differentiate it first?

many thanks!

You need to use the fact that $\frac{dy}{dx}=0$ at minima.

You do need to differentiate the equation as this gives us a gradient function. So at a minimum point the gradient is 0. So you differentiate the equation and set it equal to 0. You then solve by factorising the cubic equation. And solvin for x. To see if it is a maximum point or minimum point (or inflection point) you then take the second derivative of the equation - by differentiating the derivative. Then plug in the x values you got for dy/dx = 0 and if the second derivative is greater than 0 then it is a min point at that x value. If the second derivative is less than 0 then it is a max point at that x value. If the second derivative is 0 then you have a point of inflection.

For this question, you most definitely do NOT have to differentiate, and I would suspect you're not supposed to.

Firstly, if you really mean $y=x^4 -8^2 +3$ I would hope that it is obvious this is minimized when x = 0.

If (as seems likely) you mean $y=x^4 - 8x^2 + 3$, then treat it as a quadratic in x^2 and complete the square: $y = (x^2 - 4)^2 -13$, which is obviously mimimized when x^2 = 2, i.e. x = +/- sqrt(2).
.

Do you not mean whenever x^2 = 4, i.e. x = +/- sqrt(4)?

If the second derivative is 0 then you have a point of inflection.

Not necessarily. I always mean to write this up systematically.

1) For a "standard" x-y graph, If $dy/dx = 0$ then the gradient at that value of x is horizontal which you identify as a maximum , minimum or point of inflection. More accurately, the graph could be horizontal everywhere e.g. y = 2. A point of inflection can be sloping e.g. tan x for x = 0. Another example, between the maximum and minimum of a simple cubic, y = (x+1)(x-1)(x+2), the (sloping) point of inflection occurs where the derivative has a maximum or minimum (that's where the second derivative comes in).

For y = $x^3$ at x= 0 then dy/dx=0 and $d^2y/dx^2$ = 0 and we do have P.O.I at x=0.

Problem occurs for y= $x^4$. Here dy/dx and $d^2y/dx^2$ are zero for x=0 but NOT a P.O.I.

I think an argument goes like this.

Consider y = $x^2(x+a)^2$. For a = 0 the first derivative has an odd number of coincident (repeated) roots at x= 0 , the second derivative is zero but there is NOT a P.O.I at x=0. For a $\neq$0 you can't create an even number of coincident roots for first derivative so can't create horizontal P.O.I for this function.

Consider y = $x^3(x+a)$ . For a $\neq$0 the first derivative has an even number of repeated roots at x = 0, and there is a horizontal P.O.I at x=0. As a approaches zero the derivative gets an odd number of repeated roots at x=0 and P.O.I becomes a minimum (no longer a P.O.I even though second derivative is zero here (at x = 0)).

I am sure I can write that more clearly but I think the idea is (mostly) correct. I am sure someone will put me right if not. Whatever, this is why the textbooks avoid discussing nature of turning points when first and second derivative are zero for same value of x.

The "old boys" were taught a rule of repeated differentiation (fourth, 5th, etc derivatives) to determine the "real nature" of every turning point and the location of sloping P.O.I.s.

For this question, you most definitely do NOT have to differentiate, and I would suspect you're not supposed to.

Firstly, if you really mean $y=x^4 -8^2 +3$ I would hope that it is obvious this is minimized when x = 0.

If (as seems likely) you mean $y=x^4 - 8x^2 + 3$, then treat it as a quadratic in x^2 and complete the square: $y = (x^2 - 4)^2 -13$, which is obviously mimimized when x^2 = 4, i.e. x = +/- 2.

No need to differentiate, verifiy that the points where dy/dx = 0 are minimums as opposed to maximums etc.

Edit: fixed stupid calculation error

Wrong. You have to differentiate to find local max or local min points. This is maths 101.

Wrong. You have to differentiate to find local max or local min points. This is maths 101.

In a nutshell for higher powers you use the nth derivative test to check for local max, min and POI.
Two conditions for a POI :
1) f''(x) =0
2) f'''(x) =\ 0.
If both the 2nd and 3rd derivative =0 then use nth derivative test.

how to find the minimum points of y=x^4-8^2+3 ?
do i need to differentiate it first?

many thanks!

That's it ! Muchly thankly TeeEm.
IS what I wrote correct (if rather verbose)? I need reassurance on this.

Look up use of completing the square of a quadratic function ("actual" or "hidden"to find turning points, my friend. You will then understand DF's method.

clive

Hmm okay thank you, I will !

Wrong. You have to differentiate to find local max or local min points. This is maths 101.