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Yeah like that. What else do I have to know?do u mean stuff like stationary points on the graph are x intercepts on the graph of the derivative?
The slope of f(x) is getting less steep/levelling out if f'(x) is positive, reaching a stationary point where f'(x) is 0. The function is still going up, but not by as much. Like, I don't know, a rollercoaster reaching the top of a hill or something.Well here are some queries.
If the slope of the graph of f'(x) is negative, what about the slope of f(x)
At f'(x) = 0, there is a stationary point on the graph of f(x) where it is parallel to the x-axis, not increasing or decreasing.What does the x-intercept of f'(x) indicate?
The gradient of the gradient is 0. I think there's something about drawing a tangent here? Well, it's like a straight part of the function, but not [usually] in a nice horrizontal/vertical direction.What does the stationary point of f'(x) indicate?
The function f(x) is going up.If the graph of f'(x) is above the x-axis, what does it indicate?
Graph is concaving down. f''(x)<0 = the slope of f'(x) <0Well here are some queries.
If the slope of the graph of f'(x) is negative, what about the slope of f(x)
Stationary Pts of f(x). When f'(x) = 0, there is a turning pt. If it is a decreasing from right to left, then there is a gradient at right and negative gradient at left.What does the x-intercept of f'(x) indicate?
A pt of inflexion. When f''(x) = 0 a pt of inflexion existWhat does the stationary point of f'(x) indicate?
Gradient above 0.If the graph of f'(x) is above the x-axis, what does it indicate?