If y = -sinx + cosx, what is dy/dx? Note that it is incorrect to say "d/dx = ..."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?
Why would you think this?pip_beard said:because surly the x co-ordinate is 0?
Mark44 said:Why would you think this?
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.pip_beard said:because stationary points lie on the x axis??
No, dy/dx = -cosx - sinxpip_beard said:so therefore:
dy/dx=cosx+sinx.
pip_beard said:stationary points when diff = 0
so cosx+sinx=0 where do i go from here??
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.
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.
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.
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.
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.