Stationary points of y=-sinx+cosx

[pip_beard]

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?

 

[boboYO]

cosx+sinx=0
cosx=-sinx
1=-tanx??

 

[pip_beard]

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?

 

[Mark44]

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.

 

[Mark44]

because surly the x co-ordinate is 0?

Why would you think this?

 

[pip_beard]

so therefore:

dy/dx=cosx+sinx.

stationary points when diff = 0

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

 

[pip_beard]

Why would you think this?

because stationary points lie on the x axis??

 

[Mark44]

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.

 

[Mark44]

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?

stationary points when diff = 0

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