# Finding the Stationary Point of a Curve: y = 4 - 2/ (2x-1)^2 -x

• ibysaiyan
In summary, the conversation is about finding the derivative and stationary points of a given curve. The first question involves finding dy/dx of the curve y= 4 - 2/(2x-1)^2 -x and using the equation du/dx x dy/du to solve for the stationary points. The second question involves finding dy/dx and stationary points of the curve y= (3x-2)^3 -5x^2. The person asking the question makes mistakes in their attempts but eventually solves the problem.
ibysaiyan

## Homework Statement

Hi all , again i got stuck onto this iffy bit , question is:
find dy/dx of the curve and hence its coordinates of the statinonary point.
y= 4 - 2/(2x-1)^2 -x

du/dx x dy/du

## The Attempt at a Solution

well this is how i started off:
let u = (2x-1) , y= 4-2(u)^-2 - x
du/dx= 2;dy/du:4u^-3-1

2x[4u^(-3)-1]
[8u^(-3) -2]
to find stationary points i will let dy/dx= 0 and solve the equation ?am i on the right track so far?

oo i think i have made a mistake.. i should have multiplied gradient u (2) with the first term for the gradient of dy/du ?
EDIT:sorry for being such a spammer,i managed to solve this question but grr.. i am stuck on the next one.
question is:
curve y = (3x-2)^3 - 5x^2
find dy/dx and stationary points.

My attempt :
dy/dx= 9u^2 - 10x
we know u= (3x-2), so equation becomes
9(3x-2)^2- 10x
81x^2 -108x +26= 0
divide by 9
9x^2 -12x -2.8 ( am i over-complicating it?)

Last edited:
ibysaiyan said:
oo i think i have made a mistake.. i should have multiplied gradient u (2) with the first term for the gradient of dy/du ?
EDIT:sorry for being such a spammer,i managed to solve this question but grr.. i am stuck on the next one.
question is:
curve y = (3x-2)^3 - 5x^2
find dy/dx and stationary points.

My attempt :
dy/dx= 9u^2 - 10x
we know u= (3x-2), so equation becomes
9(3x-2)^2- 10x
81x^2 -108x +26= 0
Check your arithmetic! 9(9x^2- 12x+ 4)- 10x= 81x^2- 108x+ 36- 10x.

divide by 9
9x^2 -12x -2.8 ( am i over-complicating it?)

## 1. What is a stationary point?

A stationary point, also known as a critical point, is a point on a curve where the slope (or derivative) of the curve is zero. This means that the tangent line to the curve at that point is horizontal.

## 2. How do you find the stationary points of a curve?

To find the stationary points of a curve, you must first find the derivative of the curve. Then, set the derivative equal to zero and solve for the x-value(s). These x-values will be the coordinates of the stationary points.

## 3. What is the derivative of the given curve?

The derivative of the given curve, y = 4 - 2/ (2x-1)^2 -x, is y' = 4(2x-1)^-3 - 1. This can be simplified to y' = -8/(2x-1)^3 - 1.

## 4. How do you interpret the stationary points of a curve?

The stationary points of a curve can be interpreted as the points where the curve changes direction, from increasing to decreasing or vice versa. They can also represent the maximum or minimum values of the curve.

## 5. How many stationary points can a curve have?

A curve can have multiple stationary points, or it can have none. It is also possible for a curve to have a single stationary point that represents both a maximum and a minimum value.

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