AP Calculus AB and BC glossary

Horizontal Tangent

Also called: Horizontal tangent line

A horizontal tangent is a tangent line of slope 0. It occurs at an input c where f'(c) = 0, and the line itself is y = f(c). Every horizontal tangent at an interior point sits at a critical point, but not every critical point has one, and a flat tangent is not automatically a maximum or minimum.

f'(c) = 0

The standing counterexample is $f(x) = x^3$ . At the origin $f'(0) = 0$ , the tangent line is $y = 0$ , and the curve flattens for an instant before continuing upward. The tangent even passes through the curve rather than resting against it, which is allowed: tangency is a statement about slope, not about staying on one side.

Critical points arrive two ways and only one of them produces a flat tangent. For $f(x) = |x|$ the origin is critical because $f'$ does not exist there, and $f(x) = x^{1/3}$ has a vertical tangent at $0$ for the same reason. Solving $f'(x) = 0$ finds the horizontal tangents; the points where $f'$ is undefined have to be hunted separately.

The interior qualifier is not decoration. A critical point has to be interior to the domain, so on $f(x) = x^2$ restricted to $[0, 2]$ the one sided derivative at $x = 0$ is $0$ and the tangent there is the horizontal line $y = 0$ , yet $x = 0$ is an endpoint and not a critical point. Restricted domains are routine in the candidates test, where endpoints get checked on their own footing.

The mistake

Reporting the tangent line as $y = 0$ . The slope is zero, but the line is $y = f(c)$ . For $f(x) = x^2 - 4x + 7$ the derivative vanishes at $x = 2$ , so the horizontal tangent is $y = 3$ . Answering with the $x$ value alone, or with $y = 0$ , loses the point.

Appears in: Unit 5: Analytical Applications