What the sign of f′ tells you about f, and the one case that catches almost everybody.
Two graphs, one x. The curve is on top and its gradient function is underneath. Move the line and watch them agree: where the bottom graph is above zero the top curve climbs, where it is below zero the top curve falls.
Press play, then switch to x³ and watch what the gradient does at zero.
Those first two are reliable and worth marks. The third one is where the trouble starts, because "flat" is not the same as "turning".
Switch the toggle to f(x) = x³ and put the line at x = 0. The gradient is exactly 0, so the tangent is horizontal. But f′(x) = 3x², which is positive on both sides.
So the curve was climbing before, is flat for an instant, and climbs again. It is not a maximum and it is not a minimum. It is a stationary point, and that is all you can safely call it from f′ = 0 alone.
This is why the correct move is never "f′ = 0, therefore maximum". It is always: solve f′(x) = 0, then check the sign of f′ on each side.
For f(x) = x³ − 3x we have f′(x) = 3x² − 3 = 3(x² − 1), which is zero at x = −1 and x = 1.
| x | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| f′(x) | 9 | 0 | −3 | 0 | 9 |
| f is | up | flat | down | flat | up |
Positive, then zero, then negative: the curve climbs, flattens and falls, so x = −1 is a local maximum. Negative, zero, positive at x = 1, so that is a local minimum.
Solving f′(x) = 0 gives you the x values only. To get the point you substitute back into f, never into f′. Putting x = −1 into f′ gives 0, which is the gradient you already knew, not a coordinate.
Local, not global. The maximum here is 2, yet f(3) = 18. A local maximum is only the highest point in its own neighbourhood. If a question asks for the greatest value on an interval, check the ends of the interval as well as the stationary points.
1. A function has f′(x) = 3x² − 3. On which intervals is f increasing?
2. For some function g, g′(2) = 0, and g′ is positive just below 2 and positive just above 2. What is happening at x = 2?
3. For f(x) = x³ − 3x, give the y coordinate of the local minimum.
Writing the intervals properly matters. "f is increasing for x < −1 and x > 1" earns the mark; "f is increasing when f′ is positive" restates the definition and does not.
Solve f′(x) = 0, then substitute into f for the y value. Giving only the x values is the most common way to lose half the marks on a stationary point question.
If you are asked to justify that a point is a maximum, a sign check either side is a complete answer at this level. Saying "it looks like a maximum on my calculator" is not.
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