Finding the coordinates of stationary points when dy/dx is non zero?

Question

For instance, if we have an equation of x^3 + 3x +2, its dy/dx is 3x^2 +3 which is always positive. How do you equate it to zero and find the stationary points as most would? I think if we take the second differential, it is equals to zero at x=0, so it is a point of inflection. So does this mean that if dy/dx cannot tell us the coordinates of the stationary point, d^2y/dx^2 can tell us (because I always thought the second differential is just used for finding out what kinds of stationary point is present)?

Answer 1

There are no stationary points.

The second differential tells you what the rate of change of $dy/dx$ is, not the rate of change of $y$.

You can see here that we have no stationary points; at no point is the gradient/slope of $y$ equal to $0$.

However, at $(0,2)$ you can see the slope changing from decreasing (curving down) to increasing (curving up).

The second graph is the graph for $dy/dx$ and the turning point is when $x=0$ (slope of $dy/dx$ not $y$ here = 0); this second graph is also going through $3$ on the horizontal axis: $dy/dx=3$ not $0$ at $x=0$.

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In other words, at $x=0$ the function's slope is changing with a rate of $0$. You can visualise a straight line passing through that point and it'll make more sense:

Answer 2

Remember the definition of a stationary point.

A stationary point (aka turning point, critical point) for a function like this is a point where the first derivative is zero. That's all there is to it.

You are right that the first derivative cannot tell us stationary points here, because in fact, there are none. If you look at a graph of this function, it's always increasing and never "levels off".

You are also right that the second derivative is zero at certain points. However, at these points, the first derivative is still positive—the concavity changes, so it is a point of inflection, but it is not a stationary point.

(You might find it useful to plot this graph in Wolfram|Alpha. Also consider the graph of $arcsin(x)$. It's concave down for negative $x$, and concave up for positive, but it doesn't have any critical points either.)

Does this help?