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At the exact peak of a roller coaster, right before you plunge down the drop, you stop climbing for a fraction of a second. In that single frozen moment, your vertical speed is zero and your path is completely flat.
In calculus, the instantaneous slope of a function f(x) at any point is given by its first derivative, f′(x).
A stationary point is a coordinate (x,y) on a curve where the tangent line is completely horizontal, meaning f′(x)=0.
A broader concept is a critical value, which is any input c in the function's domain where either f′(c)=0 or f′(c) is undefined.
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Back in 1636, French mathematician Pierre de Fermat pioneered this exact technique before modern calculus even existed, noting that tangent lines flatten out at the crest of any smooth curve.
So how do we hunt down these flat spots when dealing with smooth polynomial curves?
Stationary Points on Polynomial Curves
Because polynomial curves are smooth and unbroken everywhere, their derivatives are defined for all real numbers. That means every critical value on a polynomial is guaranteed to be a stationary point.
Let's locate all stationary points for the cubic polynomial f(x)=2x3−3x2−12x+5.
First, take the derivative using the power rule: f′(x)=6x2−6x−12.
Next, set f′(x)=0 to find where the tangent slope flattens: 6(x2−x−2)=0, which factors into 6(x−2)(x+1)=0.
This gives two critical values: x=2 and x=−1. To find the actual stationary points, plug each x-value back into the original function f(x), yielding (−1,12) and (2,−15).