Tangent Line to Convex Graph

Theorem
Let $f$ be a real function that is continuous on some closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable and concave up on some open interval $\left({a \,.\,.\, b}\right)$.

Then all the tangent lines to $f$ are below the graph of $f$.

Proof

 * Concaveup.png

Let $\mathcal T$ be the tangent line to $f$ at some point $\left(c, f\left(c\right)\right)$, $c \in \left({a \,.\,.\, b}\right)$.

From the gradient-intercept form of a line, given any point $\left({x_1, y_1}\right)$ and the gradient $m$, the equation of such a line is:


 * $y - y_1 = m \left({x - x_1}\right)$

$\implies y = m\left(x - x_1\right) + y_1$

For $\mathcal T$:


 * $y = \mathcal T \left(x\right)$


 * $y_1 = f\left(c\right)$


 * $m = f^\prime \left(c\right)$


 * $x = x$


 * $x_1 = c$

so


 * $\mathcal T \left(x\right) = f^\prime \left(c\right)\left(x - c\right) + f\left(c\right)$

Consider the graph of $f$ to the right of $\left(c, f\left(c\right)\right)$, that is, any $x$ in $\left({c \,.\,.\, b}\right)$.

Let $d$ be the directed vertical distance from $\mathcal T$ to the graph of $f$. That is, if $f$ is above $\mathcal T$ then $d > 0$. If $f$ is below $\mathcal T$, then $d < 0$.

(From the diagram, it is apparent that $\mathcal T$ is below $f$, but we shall prove it analytically.)

$d$ can be evaluated by

$d = f\left(x\right) - \mathcal T \left(x\right)$


 * $= f\left(x\right) - f^\prime \left(c\right)\left(x - c\right) - f\left(c\right)$


 * $= f\left(x\right) - f\left(c\right) - f^\prime \left(c\right)\left(x - c\right)$

By the Mean Value Theorem, there exists some constant $k$ in $\left({c \,.\,.\, b}\right)$ such that


 * $f^\prime\left(k\right) = \dfrac {f\left(x\right)-f\left(c\right)}{x-c}$


 * $\implies f^\prime\left(k\right)\left(x-c\right) = f\left(x\right)-f\left(c\right)$

Substitute this into the formula for $d$


 * $d = f^\prime\left(k\right)\left(x-c\right) - f^\prime \left(c\right)\left(x - c\right)$


 * $= \left(f^\prime\left(k\right) - f^\prime \left(c\right)\right) \left(x - c\right)$

Recall that $x$ lies in the interval $\left(c..b\right)$. So $x > c$, and the quantity $x - c$ is positive.

$k$ is also in the interval $\left(c..b\right)$ and so $k > c$. By construction, $f$ is concave up.

By the definition of concavity,


 * $k > c \implies f^\prime\left(k\right) > f^\prime\left(c\right)$

which means that the quantity $\left(f^\prime\left(k\right) - f^\prime \left(c\right)\right)$ is also positive.

Then $d$ is the product of two positive quantities and is itself positive.

Similarly, consider the graph of $f$ to the left of $\left(c, f\left(c\right)\right)$, that is, any $x$ in $\left({a \,.\,.\, c}\right)$.

By the same process as above, we will have


 * $d = \left(f^\prime\left(k\right) - f^\prime \left(c\right)\right) \left(x - c\right)$

This time, $x < c$ and the quantity $x - c$ is negative.

Further, $k < c$, and so by a similar argument as above,


 * $k < c \implies f^\prime\left(k\right) < f^\prime\left(c\right)$

and the quantity $\left(f^\prime\left(k\right) - f^\prime \left(c\right)\right)$ is also negative.

Thus $d$ will be the product of two negative quantities, and will again be positive.

Also see

 * Tangent Line to Concave Down Graph
 * Tangent Line at Inflection Point