[Welz143]

the question says "describe the stationary points on the graphs of the following functions"

f(x)= x^2 + 1/x^2

how would you do this?

[Damask-]

You need to get it into a form which you can differentiate and then solve dy/dx = 0.

[ghostwalker]

You'd also need to ascertain whether they are maxima, minima, or points of inflection.

[ztibor]

(Original post by Welz143)
the question says "describe the stationary points on the graphs of the following functions"

f(x)= x^2 + 1/x^2

how would you do this?

Differentiate or draw a graph

[Khadim uz Zahra]

(Original post by Welz143)
the question says "describe the stationary points on the graphs of the following functions"

f(x)= x^2 + 1/x^2

how would you do this?

You would have to find an expression for dy/dx and, to describe the nature of the points, also find (d^2)y/dx^2 (the second differentiation).

dy/dx = 2x - 2x^-3

Then, since dy/dx is the gradient of the graph at a particular point, and as stationary points have gradient 0, then, this means at a stationary point, dy/dx = 0. Therefore, the equate dy/dx to 0 and find x, to give the values of x at which the stationary points of the graph are.

0 = 2x - 2x^-3
2x^-3 = 2x
2x^-4 = 2 (divide both sides by x)
x^-4 = 1 (divide both sides by 2)
1/x^4 = 1
x^4 = 1
x = 1

Therefore, the graph has only one stationary point and this when x = 1. Now, substitute this value of x into the equation to find the value of y when x = 1

y = (1^2) + (1/1^2)
y = 1 + 1
y = 2

Stationary point = (1,2)

Then, if they ask you to describe the nature of the stationary point, you find the second differentiation, which is simple found by differentiating the first differentiation (dy/dx)

dy/dx = 2x - 2x^-3

Therefore:
(d^2)y/dx^2 = 2 + 6x^-4

Then, find the value of (d^2)y/dx^2 at the stationary point - that is, when x = 1

(d^2)y/dx^2 = 2 + 6/1^4 = 2 + 6 = 8

The rule is that if (d^2)y/dx^2 is greater than zero, than the stationary point is a minimum and if it is less than zero, it is a maximum.

In this case, since the value is 8, which is greater than zero, then the point is a minimum.

Hope I have been helpful and clear in my explanation.