Graphs of Functions: Very Difficult Problems with Solutions

Solution:
A common point has an abscissa [tex]x\ne 0[/tex] with [tex]f(x)=g(x)[/tex].
Case [tex]x\lt 0[/tex]. Here
[tex]f(x)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\gt 0,\qquad g(x)=\frac{1}{x}\lt 0[/tex]
so the parabola lies above the axis [tex]Ox[/tex] and the left branch of the hyperbola lies below it: there are no common points.
Case [tex]x\gt 0[/tex]. Consider the function
[tex]h(x)=f(x)-g(x)=x^2+x+1-\frac{1}{x}[/tex]
On [tex]\left(0, +\infty\right)[/tex] the functions [tex]x^2[/tex], [tex]x[/tex] and [tex]-\frac{1}{x}[/tex] are all increasing, so [tex]h[/tex] is strictly increasing and has at most one zero there. (Graphically: the parabola rises while this branch of the hyperbola falls, so they cross at most once.)
Existence of a zero:
[tex]h\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{2}+1-2=-\frac{1}{4}\lt 0,\qquad h(1)=1+1+1-1=2\gt 0[/tex]
The function [tex]h[/tex] is continuous on [tex]\left[\frac{1}{2}, 1\right][/tex], so it vanishes at some point [tex]x_0[/tex] with [tex]\frac{1}{2}\lt x_0\lt 1[/tex] (equivalently, [tex]x_0[/tex] is the root of [tex]x^3+x^2+x-1=0[/tex] in this interval).
Therefore the two graphs have exactly one common point, [tex]\left(x_0, \frac{1}{x_0}\right)[/tex], and it lies in the first quadrant.

Solution:
The expressions under the absolute values vanish at the points [tex]0\lt a\lt b\lt a+b[/tex]. For [tex]0\le x\le a+b[/tex]
[tex]g(x)=x+(a+b-x)=a+b[/tex]
so the graph of [tex]g[/tex] is a horizontal segment there. For [tex]f[/tex]:
[tex]0\le x\le a:\quad f(x)=(a-x)+(b-x)=a+b-2x[/tex]
[tex]a\le x\le b:\quad f(x)=(x-a)+(b-x)=b-a[/tex]
[tex]b\le x\le a+b:\quad f(x)=(x-a)+(x-b)=2x-a-b[/tex]
The graph of [tex]f[/tex] goes down from [tex](0, a+b)[/tex] to [tex](a, b-a)[/tex], stays horizontal up to [tex](b, b-a)[/tex] and rises to [tex](a+b, a+b)[/tex]. So on [tex]\left[0, a+b\right][/tex] the graphs meet only at the endpoints, and [tex]f\lt g[/tex] in between.
The enclosed figure is bounded above by the segment [tex]y=a+b[/tex] and below by this broken line. It is an isosceles trapezoid with bases [tex]a+b[/tex] and [tex]b-a[/tex] and height [tex](a+b)-(b-a)=2a[/tex]:
[tex]S=\frac{(a+b)+(b-a)}{2}\times 2a=b\times 2a=2ab[/tex]
Check by parts: two triangles with area [tex]\frac{1}{2}\times a\times 2a=a^2[/tex] each and a rectangle with area [tex](b-a)\times 2a[/tex]:
[tex]2a^2+2a(b-a)=2a^2+2ab-2a^2=2ab[/tex]

Solution:
The solutions are the abscissas of the common points of the sine curve [tex]y=\sin x[/tex] and the line [tex]y=\frac{x}{13}[/tex]. Both functions are odd, so the solutions are symmetric about [tex]0[/tex]: [tex]x=0[/tex] is a solution, and it suffices to count the positive solutions and double their number.
Since [tex]\left|\sin x\right|\le 1[/tex], every solution satisfies [tex]\left|x\right|\le 13[/tex]. For [tex]x\gt 0[/tex] on the intervals [tex]\left[\pi, 2\pi\right][/tex] and [tex]\left[3\pi, 4\pi\right][/tex] we have [tex]\sin x\le 0\lt\frac{x}{13}[/tex], so there are no solutions there. Note that [tex]4\pi\lt 13\lt\frac{9\pi}{2}[/tex], because [tex]\frac{26}{9}\lt\pi\lt\frac{13}{4}[/tex].
Let [tex]h(x)=\sin x-\frac{x}{13}[/tex]. Where [tex]\sin x\gt 0[/tex], we have [tex]h''(x)=-\sin x\lt 0[/tex], so [tex]h[/tex] is concave there and has at most two zeros on each such interval (a line meets each arch of the sine curve in at most two points).
On [tex]\left[0, \pi\right][/tex]:
[tex]h(0)=0,\qquad h'(0)=1-\frac{1}{13}\gt 0,\qquad h(\pi)=-\frac{\pi}{13}\lt 0[/tex]
The sine curve starts above the line and ends below it, so besides [tex]x=0[/tex] there is a zero in [tex]\left(0, \pi\right)[/tex], and since [tex]h[/tex] has at most two zeros on [tex]\left[0, \pi\right][/tex], there is exactly one solution in [tex]\left(0, \pi\right)[/tex].
On [tex]\left[2\pi, 3\pi\right][/tex] (using [tex]5\pi\lt 26[/tex]):
[tex]h(2\pi)=-\frac{2\pi}{13}\lt 0,\qquad h\left(\frac{5\pi}{2}\right)=1-\frac{5\pi}{26}\gt 0,\qquad h(3\pi)=-\frac{3\pi}{13}\lt 0[/tex]
so there is a zero in each of [tex]\left(2\pi, \frac{5\pi}{2}\right)[/tex] and [tex]\left(\frac{5\pi}{2}, 3\pi\right)[/tex], and at most two in total: exactly two solutions.
On [tex]\left[4\pi, 13\right][/tex], using [tex]\sin t\le t[/tex] for [tex]t\ge 0[/tex] and [tex]\frac{25}{8}\lt\pi[/tex]:
[tex]\sin x=\sin (x-4\pi)\le x-4\pi\le 13-4\pi\lt\frac{1}{2}\lt\frac{12}{13}\lt\frac{4\pi}{13}\le\frac{x}{13}[/tex]
so there are no solutions there. Thus there are [tex]1+2=3[/tex] positive solutions, [tex]3[/tex] negative ones and [tex]x=0[/tex]:
[tex]3+3+1=7[/tex]