Stationary points of y=-sinx+cosx

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In summary, to find the stationary points of y = -sinx + cosx for the domain -pi < x < pi, we first differentiate the function to get dy/dx = -cosx - sinx. Then, we set this equal to zero and solve for x, which gives us the equation 1 + tanx = 0. From here, we can solve for x to get the two stationary points of (-pi/4, -√2) and (3pi/4, √2).
  • #1
pip_beard
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Homework Statement


Solve the stationary points of y=-sinx+cosx for domain -pi<x<pi


Homework Equations





The Attempt at a Solution


Differentiate: d/dx=cosx+sinx But how do i solve?
 
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  • #2
cosx+sinx=0
cosx=-sinx
1=-tanx??
 
  • #3
ive got the answers as.. (-pi/4,-rt2) and (3pi/4, rt2) Are the answers wrong?

I don't know how they got these??

because surly the x co-ordinate is 0?
 
  • #4
pip_beard said:

Homework Statement


Solve the stationary points of y=-sinx+cosx for domain -pi<x<pi


Homework Equations





The Attempt at a Solution


Differentiate: d/dx=cosx+sinx But how do i solve?
If y = -sinx + cosx, what is dy/dx? Note that it is incorrect to say "d/dx = ..."

If you meant dy/dx = ..., you have made a mistake. Try again.

Also, your notation is not correct. d/dx is an operator that is written to the left of some function. In contrast, dy/dx is the derivative of y with respect to x.
 
  • #5
pip_beard said:
because surly the x co-ordinate is 0?
Why would you think this?
 
  • #6
so therefore:

dy/dx=cosx+sinx.

stationary points when diff = 0

so cosx+sinx=0 where do i go from here??
 
  • #7
Mark44 said:
Why would you think this?

because stationary points lie on the x axis??
 
  • #8
pip_beard said:
because stationary points lie on the x axis??
x-values lie on the x-axis, but stationary points lie on the curve, which might not even touch the x-axis. For example, the only stationary point on the graph of y = x^2 + 1 is at (0, 1). This is not a point on the x-axis.
 
  • #9
pip_beard said:
so therefore:

dy/dx=cosx+sinx.
No, dy/dx = -cosx - sinx

To find the stationary points, set dy/dx to zero.
-cosx - sinx = 0
==> cosx + sinx = 0
==> 1 + tanx = 0 (dividing both sides by cosx)
Can you continue?

pip_beard said:
stationary points when diff = 0

so cosx+sinx=0 where do i go from here??
 

1. What is the definition of stationary points?

Stationary points are points on a graph where the gradient is equal to zero, meaning there is no change in the value of the function at that point.

2. How do you find the stationary points of y=-sinx+cosx?

To find the stationary points of y=-sinx+cosx, you need to take the derivative of the function and set it equal to zero. This will give you the x-values of the stationary points. You can then plug these x-values back into the original function to find the corresponding y-values.

3. What is the significance of stationary points in a graph?

Stationary points can give us important information about the behavior of a function. They can indicate the maximum or minimum values of a function, as well as points of inflection or points where the function is constant.

4. Can a function have more than one stationary point?

Yes, a function can have multiple stationary points. This can occur when the function has multiple peaks or valleys, or when there is a point of inflection where the gradient is equal to zero.

5. How do stationary points relate to the overall shape of a graph?

The number and location of stationary points can greatly affect the overall shape of a graph. For example, if a function has no stationary points, it will be a straight line. If a function has one stationary point, it will have either a maximum or minimum point. And if a function has multiple stationary points, it can have multiple peaks and valleys or a more complex shape.

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