Extra credit

In the February 2014 issue of the Mathematics Magazine the following is stated (Connie Xu, Montgomery Blair High School):

Let $f(x)$ be a twice-differentiable function such that $f''(x)$ is continuous and positive for all $x$. Then the following are equivalent:

Suppose for any $a < b$ the tangent lines at $a$ and $b$ intersect at $x=c$. The area between $f(x)$ and the tangent line at $a$ between $a$ and $c$ is equal to that area between $f(x)$ and the tangent lines at $b$ between $c$ and $b$.

Furthermore, the sum of these two described areas depends on $b-a$ and not the values of $a$ and $b$.

You have the following assignment:

For the function $f(x) = x^2$ verify numerically that the second part above holds for at least 4 pairs of different values of $a$ and $b$ with $b-a$ being a constant for all 4 pairs.

For some other function $f(x)$ satisfying the assumptions, show that you can find a pair $a$ and $b$ for which the second part does not hold (the two areas are different.)

To get you started, the following shows how to plot the area between a function and the tangent lines:

using MTH229
plotly()

f(x) = x^2
a,b = -0.5, 1
plot([f, tangent(f,a), tangent(f,b)], a, b)

And this shows how to find the intersection point between two tangent lines:

g(x) = tangent(f, a)(x) - tangent(f,b)(x)    # g(x) = 0 is when two tangent lines intersect
c = fzero(g, a,b)

0.25

Finally the area between a function and the tangent line is found with

h(x) =  f(x) - tangent(f, a)(x)
area, error = quadgk(h, a, b) # area between a and b is for example only

(1.1249999999999998,0.0)

(This works as $f''(x) > 0$ implies that function is concave up, so that the tangent line is below the function.)

Okay, give it a go here: