Graphs of Functions: Difficult Problems with Solutions

Solution:
The area condition gives [tex]xy=400[/tex], i.e. [tex]y=\frac{400}{x}[/tex] with [tex]x\gt 0[/tex]. The cut rectangle must fit inside the card with its sides parallel to the sides of the card, and it can be placed in two ways.
If the side [tex]x[/tex] lies along the side of length [tex]20[/tex], then [tex]x\le 20[/tex] and [tex]y\le 40[/tex]:
[tex]x\le 20,\quad \frac{400}{x}\le 40\iff x\ge 10\qquad\Rightarrow\qquad 10\le x\le 20[/tex]
If the side [tex]x[/tex] lies along the side of length [tex]40[/tex], then [tex]x\le 40[/tex] and [tex]y\le 20[/tex]:
[tex]x\le 40,\quad \frac{400}{x}\le 20\iff x\ge 20\qquad\Rightarrow\qquad 20\le x\le 40[/tex]
Every [tex]x[/tex] from either case is possible, so the domain is the union of the two intervals:
[tex]\left[10, 20\right]\cup\left[20, 40\right]=\left[10, 40\right][/tex]
The graph is the arc of the hyperbola [tex]y=\frac{400}{x}[/tex] from [tex](10, 40)[/tex] to [tex](40, 10)[/tex]. (Considering only one placement gives [tex]\left[10, 20\right][/tex] or [tex]\left[20, 40\right][/tex]; ignoring the condition on [tex]y[/tex] gives [tex]\left(0, 40\right][/tex].)

Solution:
Draw the graphs of [tex]y_1=\left|x\right|[/tex] (an angle with vertex [tex](0, 0)[/tex]) and [tex]y_2=\left|x+1\right|[/tex] (the same angle shifted [tex]1[/tex] unit to the left, with vertex [tex](-1, 0)[/tex]). They are symmetric with respect to the line [tex]x=-\frac{1}{2}[/tex] and meet only where the decreasing branch of [tex]y_1[/tex] meets the increasing branch of [tex]y_2[/tex]:
[tex]-x=x+1\Rightarrow x=-\frac{1}{2},\quad y=\frac{1}{2}[/tex]
To the left of the intersection point [tex]\left(-\frac{1}{2}, \frac{1}{2}\right)[/tex] the graph of [tex]y_1[/tex] lies above the graph of [tex]y_2[/tex] (e.g. at [tex]x=-2[/tex]: [tex]2\gt 1[/tex]), and to the right of it, below (at [tex]x=0[/tex]: [tex]0\lt 1[/tex]). The inequality is strict, so [tex]x=-\frac{1}{2}[/tex], where the values are equal, is excluded:
[tex]x\lt -\frac{1}{2}[/tex]
Algebraic check: both sides are non-negative, so we may square them:
[tex]x^2\gt x^2+2x+1\iff 2x+1\lt 0\iff x\lt -\frac{1}{2}[/tex]

Solution:
The expressions under the absolute value signs vanish at [tex]x=0[/tex] and [tex]x=1[/tex]; on each of the three intervals the function is linear:
[tex]f(x)=\left\{\begin{array}{ll}-x-(1-x)=-1, & x\lt 0\\ x-(1-x)=2x-1, & 0\le x\lt 1\\ x-(x-1)=1, & x\ge 1\end{array}\right.[/tex]
The graph consists of the horizontal ray [tex]y=-1[/tex] for [tex]x\lt 0[/tex], the segment from [tex](0, -1)[/tex] to [tex](1, 1)[/tex], and the horizontal ray [tex]y=1[/tex] for [tex]x\ge 1[/tex]. Its projection on the [tex]y[/tex]-axis is the closed interval [tex]\left[-1, 1\right][/tex]; both ends are attained:
[tex]f(0)=-1,\qquad f(1)=1[/tex]
The same bound also follows from the triangle inequality:
[tex]\left|f(x)\right|=\left|\,\left|x\right|-\left|x-1\right|\,\right|\le\left|x-(x-1)\right|=1[/tex]
Hence the range is [tex]\left[-1, 1\right][/tex].

Solution:
Compute the values at the ends of the pieces:
[tex]f(1)=3\times 1-3=0,\quad 3\times 2-3=3=f(2),\quad f(5)=10.5-7.5=3,\quad f(7)=10.5-10.5=0[/tex]
So the graph is the segment from [tex](1, 0)[/tex] to [tex](2, 3)[/tex], the horizontal segment from [tex](2, 3)[/tex] to [tex](5, 3)[/tex] and the segment from [tex](5, 3)[/tex] to [tex](7, 0)[/tex]. The pieces join, so the graph is the broken line [tex](1, 0)\to(2, 3)\to(5, 3)\to(7, 0)[/tex], lying on or above the [tex]x[/tex]-axis. Together with the segment from [tex]1[/tex] to [tex]7[/tex] on the [tex]x[/tex]-axis it bounds a trapezoid with bases [tex]7-1=6[/tex] and [tex]5-2=3[/tex] and height [tex]3[/tex]:
[tex]S=\frac{6+3}{2}\times 3=\frac{27}{2}[/tex]
Check by parts (two triangles and a rectangle):
[tex]S=\frac{1}{2}\times 1\times 3+3\times 3+\frac{1}{2}\times 2\times 3=\frac{3}{2}+9+3=\frac{27}{2}[/tex]

Solution:
The function is defined only where the denominator is non-zero, i.e. for [tex]x\ne 0[/tex]. For these [tex]x[/tex] we can cancel:
[tex]f(x)=\frac{x^3}{x^2}=x,\qquad x\ne 0[/tex]
So the graph is the line [tex]y=x[/tex] (the bisector of the first and third quadrants) without the point [tex]O(0, 0)[/tex], which is drawn as a hollow circle.
The horizontal line [tex]y=m[/tex] meets the line [tex]y=x[/tex] at the single point [tex](m, m)[/tex]. This point belongs to the graph of [tex]f[/tex] exactly when [tex]m\ne 0[/tex]. For [tex]m=0[/tex] the line [tex]y=0[/tex] passes through the removed point, and the equation [tex]f(x)=0[/tex] has no solution. Hence
[tex]m\ne 0[/tex]

Solution:
The function makes sense only for [tex]a\ne 0[/tex]. Then
[tex]y=\frac{-2a-1}{a}\times \frac{1}{x}\ \Rightarrow\ k=\frac{-2a-1}{a}=-2-\frac{1}{a}[/tex]
For an integer [tex]a[/tex], the number [tex]k[/tex] is an integer exactly when [tex]\frac{1}{a}[/tex] is an integer, i.e. for [tex]a=1[/tex] or [tex]a=-1[/tex]:
[tex]a=1\ \Rightarrow\ k=-3,\ \ y=-\frac{3}{x}\qquad\qquad a=-1\ \Rightarrow\ k=\frac{2-1}{-1}=-1,\ \ y=-\frac{1}{x}[/tex]
The hyperbola [tex]y=\frac{k}{x}[/tex] lies in the second and fourth quadrants exactly when [tex]k\lt 0[/tex] (then [tex]y\gt 0[/tex] for [tex]x\lt 0[/tex] and [tex]y\lt 0[/tex] for [tex]x\gt 0[/tex]). Both values give [tex]k\lt 0[/tex], so both are solutions:
[tex]a\in\left\{-1, 1\right\}[/tex]

Solution:
The expressions under the absolute values change sign at [tex]x=-1[/tex] and [tex]x=\frac{1}{2}[/tex]. On each interval we remove the absolute values:
[tex]x\lt -1:\quad f(x)=-(2x-1)+(x+1)+x+2=4[/tex]
[tex]-1\le x\lt\frac{1}{2}:\quad f(x)=-(2x-1)-(x+1)+x+2=2-2x[/tex]
[tex]x\ge\frac{1}{2}:\quad f(x)=(2x-1)-(x+1)+x+2=2x[/tex]
So the graph consists of the horizontal ray [tex]y=4[/tex] for [tex]x\le -1[/tex], the segment from [tex](-1, 4)[/tex] to [tex]\left(\frac{1}{2}, 1\right)[/tex] and the ray [tex]y=2x[/tex] for [tex]x\ge\frac{1}{2}[/tex]. The least value of the function is [tex]f\left(\frac{1}{2}\right)=1[/tex].
Let [tex]c=k+2[/tex] and move the horizontal line [tex]y=c[/tex]. If [tex]c\lt 1[/tex], there are no common points; if [tex]c=1[/tex], there is one point ([tex]x=\frac{1}{2}[/tex]).
If [tex]1\lt c\lt 4[/tex], there are two points: [tex]x=\frac{2-c}{2}\in\left(-1, \frac{1}{2}\right)[/tex] on the segment and [tex]x=\frac{c}{2}\gt\frac{1}{2}[/tex] on the ray.
If [tex]c=4[/tex], there are infinitely many (every [tex]x\le -1[/tex], and also [tex]x=2[/tex]); if [tex]c\gt 4[/tex], there is one point ([tex]x=\frac{c}{2}[/tex]).
Exactly two solutions occur only for [tex]1\lt c\lt 4[/tex]:
[tex]1\lt k+2\lt 4\ \Leftrightarrow\ -1\lt k\lt 2\ \Leftrightarrow\ k\in\left(-1, 2\right)[/tex]

Solution:
The line [tex]g[/tex] passes through [tex]B[/tex] and [tex]A[/tex]:
[tex]g(0)=n=\frac{16}{3},\qquad g(8)=8k+\frac{16}{3}=0\ \Rightarrow\ k=-\frac{2}{3},\qquad g(x)=-\frac{2}{3}x+\frac{16}{3}[/tex]
The function [tex]f[/tex]. If [tex]x\ge 1[/tex], then [tex]\left|1-x\right|=x-1[/tex] and [tex]3x+1-(x-1)=2x+2\gt 0[/tex], so
[tex]f(x)=2x+2-2x=2[/tex]
If [tex]x\lt 1[/tex], then [tex]\left|1-x\right|=1-x[/tex] and [tex]3x+1-(1-x)=4x[/tex], so [tex]f(x)=\left|4x\right|-2x[/tex]:
[tex]0\le x\lt 1:\ f(x)=4x-2x=2x,\qquad x\lt 0:\ f(x)=-4x-2x=-6x[/tex]
So the graph of [tex]f[/tex] is the ray [tex]y=-6x[/tex] for [tex]x\lt 0[/tex], the segment from [tex]O(0, 0)[/tex] to [tex]C(1, 2)[/tex] and the horizontal ray [tex]y=2[/tex] for [tex]x\ge 1[/tex]. Intersection points with [tex]g[/tex]:
[tex]-6x=-\frac{2}{3}x+\frac{16}{3}\ \Rightarrow\ x=-1\lt 0,\quad P(-1, 6)[/tex]
[tex]2=-\frac{2}{3}x+\frac{16}{3}\ \Rightarrow\ x=5\ge 1,\quad Q(5, 2)[/tex]
The equation [tex]2x=-\frac{2}{3}x+\frac{16}{3}[/tex] gives [tex]x=2[/tex], which is rejected, because the formula [tex]f(x)=2x[/tex] holds only for [tex]0\le x\lt 1[/tex].
The graphs meet only at [tex]P[/tex] and [tex]Q[/tex], and at [tex]x=0[/tex] we have [tex]g(0)=\frac{16}{3}\gt 0=f(0)[/tex], so for [tex]-1\lt x\lt 5[/tex] the line [tex]g[/tex] lies above the graph of [tex]f[/tex]; both functions are non-negative there. The enclosed figure has vertices [tex]P(-1, 6)[/tex], [tex]O(0, 0)[/tex], [tex]C(1, 2)[/tex], [tex]Q(5, 2)[/tex], and its area is the area under [tex]g[/tex] minus the area under [tex]f[/tex] on [tex]\left[-1, 5\right][/tex].
Under [tex]g[/tex]: a trapezoid with parallel sides [tex]g(-1)=6[/tex] and [tex]g(5)=2[/tex] and width [tex]6[/tex]. Under [tex]f[/tex]: two triangles and a rectangle:
[tex]S_g=\frac{6+2}{2}\times 6=24,\qquad S_f=\frac{1}{2}\times 1\times 6+\frac{1}{2}\times 1\times 2+4\times 2=3+1+8=12[/tex]
[tex]S=S_g-S_f=24-12=12[/tex]