Problem #1: The area of a triangle is $\dfrac{1}{2}bh$. In this case, the base and height are both determined by the $x$- and $y$-intercepts of the tangent line. In general, points on the parabola given are of the form $(u,1-u^2)$. The slope of a tangent line at the point $(u,1-u^2)$ is given by plugging $u$ into the derivative. The derivative of the parabola is $y' = -2x$, so the slope of the tangent line is $-2u$. Then, in point slope form, $y-y_0 = m(x-x_0)$. Plugging in:

$y-(1-u^2) = -2u(x-u)$

Converting to slope-intercept form:

$y = -2ux + u^2+1 = mx+b$

Since $-2ux = mx$, obviously, $u^2+1 = b$, so the $y$-intercept is $u^2+1$.

Then, what is the $x$-intercept? That is when $y=0$. Plug in for $y=0$ and solve for $x$:

$2ux = u^2+1 \Rightarrow x = \dfrac{u^2+1}{2u}$

This gives you your base and height. So, the area is:

$A = \dfrac{1}{2}bh = \dfrac{1}{2}\dfrac{u^2+1}{2u}(u^2+1) = \dfrac{1}{4}(u+u^{-1})(u^2+1) = \dfrac{1}{4}(u^3+2u+u^{-1})$

Now, to minimize that, take the derivative with respect to $u$ and set it equal to zero to find critical numbers:

$A'(u) = \dfrac{3u^2+2-u^{-2}}{4} = 0$

Multiplying out by $4u^2$ gives:

$3u^4 + 2u^2-1 = 0$

$(u^2+1)\left(u^2-\dfrac{1}{3}\right) = 0$, so $u^2 = -1$ or $u^2 = \dfrac{1}{3}$. Since $\sqrt{-1}$ is not a real number, it must be that $u = \pm \sqrt{\dfrac{1}{3}}$. Since the negative answer is not an $x$-value in the first quadrant, we reject it. Hence, $u = \sqrt{\dfrac{1}{3}}$. Next, check to see if it is truly a minimum. You can use the second derivative test (if the second derivative is positive at $u = \sqrt{\dfrac{1}{3}}$, you found a minimum).

For question 2: By the logarithm rules, $\ln|2x| = \ln 2 + \ln |x|$. Since $\ln 2 \approx 0.693$, it is a constant. Hence, just as the corollary said, $\ln|x| \text{ and } \ln|2x|$ differ by just a constant.