What exactly does this mean: "The graph of $f$ lies above all of its tangents" The Stewart calculus book defines "concave upward" as:

If the graph of $f$ lies above all of its tangents on an interval $I$, then it is called concave upward on $I$.

I don't think it means the graph of $f$ lies completely above each tangent, because the graph of $f$ must touch each tangent at a point, namely the point of tangency.
So does the definition mean that the graph of $f$ lies above or touches each tangent?
Or does it mean that the graph of $f$ lies completely above each tangent except at the point of tangency?
Or does it mean something else?
 A: Usually, "concave upward" a.k.a. convex is defined in terms of secants, not tangents, for greater generality: $f$ is convex if the graph of $f$ lies below all secant lines. Formally, for $x,y \in I$ and $0\le t \le 1$, $$f(tx + (1-t)y) \le t f(x) + (1-t)f(y).$$ This allows for the secant line to touch not only at the endpoints, but anywhere. A function is called strictly convex if whenever $x \ne y$ and $t \ne 0,1$, the inequality is strict.
For functions $f$ that have a derivative $f'$ on $I$, the definition above is equivalent to the inequality that for all $x,y \in I$, $$f(y) \ge f(x) + (y-x) f'(x)$$ which is the tangent line condition. Again, this allows for the tangent line to touch the graph at multiple points - though it's worth noting that the tangent line won't be able to leave the graph and come back. Two interesting examples:

*

*The function $f(x) = x$ is convex, even though its tangent line is itself at all points.

*The function $f(x) = \begin{cases}x^2 & x<0 \\ 0 & 0\le x \le 1 \\ (x-1)^2 & x>1\end{cases}$ is convex on the real line. Its tangent line at any point between $0$ and $1$ is the $x$-axis, which touches it along an entire interval.

The definition of strict convexity is equivalent to having $f(y) > f(x) + (y-x) f'(x)$ whenever $y \ne x$.
So here's your answer: Convex, or concave upward, allows the tangent line to keep touching the graph for an extended interval. Strictly convex requires the tangent line to lie below the graph everywhere except at the point of tangency.
Finally, for functions $f$ that have a second derivative, the definition is equivalent to having $f''(x) \ge 0$ for all $x \in I$. I mention this only to say that though $f''(x) > 0$ implies strict convexity, it is not required for strict convexity: for example, $f(x) = x^4$ is strictly convex on the real line, even though $f''(0) = 0$.

A final note: terminology always varies between sources. However, in any case where it matters, it is always better to have two terms "convex" and "strictly convex" that distinguish the two different situations, rather than just have one term for a case that might not be the case you want.
A: I think it means this.
For any $x \in I$, let the tangent line to the curve $y = f(x)$ at $x$ have the equation $t(x)$.
Then, iff $f(x_1) \geq t(x_0)$ for any two points with $x_0,x_1 \in I$ with strict inequality ($f(x_1) > t(x_0)$) being achieved for $x_0 \neq x_1$, the curve $y = f(x)$ is concave upward over the interval $I$.
