\(f(x) = \dfrac{2x - 1}{x^2}\)

\(f(x)\) is undefined at \(x=0\)

Domain: \((-\infty, 0) \cup (0, \infty)\)

clear
pkg load symbolic
syms x
f = (2*x-1)*x^(-2)

df = simplify(diff(f,x))
dff = simplify(diff(df,x))

octave> octave> f = (sym)

  2⋅x - 1
  ───────
      2
     x
octave> df = (sym)

  2⋅(1 - x)
  ─────────
       3
      x
dff = (sym)

  2⋅(2⋅x - 3)
  ───────────
        4
       x

Both \(f'(x)\) and \(f''(x)\) are undefined at \(x=0\). \(f'(x) = 0\) when \(x=1\). \(f''(x) = 0\) when \(x = 3/2\)

Let's check the function for vertical asymptotes

\(\lim_{x \to 0^-} f(x) = - \infty\)

\(\lim_{x \to 0^+} f(x) = - \infty\)

So \(x = 0\) is the vertical asymptotes. Let's check the function for horizontal asymptotes.

\(\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \dfrac{2x-1}{x^2} = \lim_{x \to \pm \infty} \dfrac{x(2 - 1/x}{x^2}\)

\(= \lim_{x \to \pm \infty} \dfrac{1}{x}(2 - \dfrac{1}{x}) = 0.2 = 0\)

So \(y = 0\) is the horizontal asymptote.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
f(x) = (2*x - 1) / (x**2)
set yrange [-5:5]
set xrange [-5:5]
set label "I" at 1.5, f(1.5) center point pointtype 7 pointsize 2 offset 1
set label "C" at 1, f(1) center point pointtype 6 pointsize 2 offset -2
plot (x > 0 ? f(x) : 1/0) ls 1 t "2x-1/x^2", (x < 0 ? f(x) : 1/0) ls 1 t "", 0 with lines ls 2 t "asymptote"

C = Critical number

I = Inflection point

\(g(x) = \dfrac{2}{x^2} = \dfrac{1}{x}\)

\(g(x)\) is undefined at \(x = 0\)

Domain: \((-\infty, 0) \cup (0 , \infty)\)

clear
pkg load symbolic
syms x
g = 2/(x^2) - 1/x
dg = simplify(diff(g,x))
dgg = simplify(diff(dg, x))

octave> octave> g = (sym)

    1   2
  - ─ + ──
    x    2
        x
dg = (sym)

  x - 4
  ─────
     3
    x
dgg = (sym)

  2⋅(6 - x)
  ─────────
       4
      x

Both \(g'(x)\) and \(g''(x)\) are undefined at \(x=0\).

\(g'(x) = 0\) when \(x = 4\)

\(g''(x) = 0\) when \(x = 6\)

Let's check the function for vertical asymptotes

\(\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} \dfrac{1}{x}(2/x - 1) = -\infty\)

\(\lim_{x \to 0^+} g(x) = \infty\)

So \(x = 0\) is a vertical asymptote.

\(\lim_{x \to \pm \infty} g(x) = \lim_{x \to \pm \infty} g(x) \dfrac{2}{x^2} - \dfrac{1}{x} = 0 - 0 = 0\)

So \(y = 0\) is the horizontal asymptote.

Inflection point: 6

Critical number: 4

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
g(x) = (2/(x**2)) - (1/x)
set yrange [-1:8]
set xrange [-5:10]
set label "I" at 6, g(6) left point pointtype 7 pointsize 2 offset 1
set label "C" at 4, g(4) center point pointtype 6 pointsize 2 offset -1
plot (x > 0 ? g(x) : 1/0) ls 1 t "2/x^2 - 1/x", (x < 0 ? g(x) : 1/0) ls 1 t "" , 0 with lines ls 2 t "asymptote"

\(f(x) = \dfrac{x}{4-x^2}\)

\(f(x)\) is undefined at \(x=2, -2\)

Domain: \((-\infty, 2) \cup (-2,2) \cup (2, \infty)\)

clear
pkg load symbolic
syms x
f = x/(4-x^2)

df = simplify(diff(f,x))
dff = simplify(diff(df,x))

octave> octave> f = (sym)

    x
  ──────
       2
  4 - x
octave> df = (sym)

     2
    x  + 4
  ─────────
          2
  ⎛ 2    ⎞
  ⎝x  - 4⎠
dff = (sym)

       ⎛ 2     ⎞
  -2⋅x⋅⎝x  + 12⎠
  ───────────────
             3
     ⎛ 2    ⎞
     ⎝x  - 4⎠

Both \(f'(x)\) and \(f''(x)\) are undefined at \(x = 2, -2\). \(f'(x) = 0\) when \(x = \emptyset\)

\(f''(x) = 0\) when \(x=0\)

Inflection point: 0

Let's check the function for vertical asymptotes:

\(\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} \dfrac{x}{x^2(4/x^2 - 1)} = \infty\)

\(\lim_{x \to -2^+} f(x) = -\infty\)

\(\lim_{x \to 2^-} f(x) = \infty\)

\(\lim_{x \to 2^+} f(x) = -\infty\)

So \(x=2\) and \(x=-2\) are the vertical asymptotes for the function. Let's check the function for horizontal asymptotes:

\(\lim_{x \to \pm \infty} f(x) = \lim_{x \tp \pm \infty} \dfrac{1}{4/x - x} = 0\)

So \(y = 0\) is the horizontal symptote

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
f(x) = (x/(4 - x**2))
set yrange [-3:3]
set xrange [-5:5]
set label "I" at 0, f(0) center point pointtype 7 pointsize 2 offset -2
plot (x < -2 ? f(x) : 1/0) ls 1 t "x/(4-x^2)", (x < 2 ? f(x) : 1/0) ls 1 t "" , (x > 2 ? f(x) : 1/0) ls 1 t "", 0 with lines ls 2 t "asymptote"

\(g(x) = \dfrac{x^2}{4-x^2}\)

\(g(x)\) is undefined at \(x=2,-2\)

Domain: \((-\infty, -2) \cup (-2,2) \cup (2, \infty)\)

clear
pkg load symbolic
syms x
g = x^2/(4-x^2)

dg = simplify(diff(g,x))
dgg = simplify(diff(dg, x))

octave> octave> g = (sym)

     2
    x
  ──────
       2
  4 - x
octave> dg = (sym)

     8⋅x
  ─────────
          2
  ⎛ 2    ⎞
  ⎝x  - 4⎠
dgg = (sym)

   ⎛    2     ⎞
  -⎝24⋅x  + 32⎠
  ──────────────
            3
    ⎛ 2    ⎞
    ⎝x  - 4⎠

Both \(g'(x)\) and \(g''(x)\) are undefined at \(x=2, -2\). \(g'(x) = 0\) when \(x=0\). \(g''(x) = 0\) when \(x = \emptyset\)

Let's check the function for vertical asymptotes:

\(\lim_{x \to -2^-} g(x) = -\infty\)

\(\lim_{x \to -2^+} g(x) = \infty\)

\(\lim_{x \to 2^-} g(x) = \infty\)

\(\lim_{x \to 2^+} g(x) = -\infty\)

So \(x = 2\) and \(x = -2\) are the vertical asymptotes.

\(\lim_{x \to \pm \infty} g(x) = \lim_{x \to \pm \infty} \dfrac{1}{4/x^2 - 1} = \dfrac{1}{0-1} = -1\)

So \(y = -1\) is the horizontal asymptote.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
f(x) = (x**2/(4 - x**2))
set yrange [-3:3]
set xrange [-5:5]
set label "C" at 0, f(0) center point pointtype 7 pointsize 2 offset -2
plot (x < -2 ? f(x) : 1/0) ls 1 t "x^2/(4-x^2)", (x < 2 ? f(x) : 1/0) ls 1 t "" , (x > 2 ? f(x) : 1/0) ls 1 t "", -1 with lines ls 2 t "asymptote"

function y = firstDerivative(x)
  num = 8 * x;
  den = (4-x^2)^2;
  y = num / den;
end

ans = [firstDerivative(-1), firstDerivative(-3), firstDerivative(1), firstDerivative(3)]

function y = secondDerivative(x)
  num = 24*(x^2) + 32
  den = (4-x^2)^3
  y = num / den;
end

ans = [secondDerivative(-3), secondDerivative(-1), secondDerivative(0), secondDerivative(1), secondDerivative(3)]

\(f(x)= \dfrac{x^3 +2}{x}\)

\(f(x)\) is undefined at \(x=0\)

Domain: \((-\infty, 0) \cup (0, \infty)\)

clear
pkg load symbolic
syms x
f = (x^3 + 2)/x

df = simplify(diff(f,x))
dff = simplify(diff(df,x))

octave> octave> f = (sym)

   3
  x  + 2
  ──────
    x
octave> df = (sym)

        2
  2⋅x - ──
         2
        x
dff = (sym)

      4
  2 + ──
       3
      x

Both \(f'(x)\) and \(f''(x)\) are undefined at \(x=0\). \(f'(x) = 0\) when \(x = 1\). \(f''(x) = 0\) when \(x = -(2)^{1/3}\)

function y = firstDerivative(x)
  num = 2*x^3 - 2
  den = x^2
  y = num / den;
end

ans = [firstDerivative(-2), firstDerivative(-1.25), firstDerivative(-1), firstDerivative(1), firstDerivative(2)]

function y = secondDerivative(x)
  num = 2*x^3 + 4
  den = x^3
  y = num / den;
end

ans = [secondDerivative(-2), secondDerivative(-1), secondDerivative(1), secondDerivative(2)]

Let's check the function for vertical asymptotes

\(\lim_{x \to 0^-} f(x) = -\infty\)

\(\lim_{x \to 0^+} f(x) = \infty\)

So \(x = 0\) is the vertical asymptotes.

\(\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \dfrac{1 + 2/x^3}{1/x^2} = \infty\)

So there is no horizontal asymptotes.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
f(x) = (x**3 + 2)/x
set yrange [-10:10]
set xrange [-5:5]
set label "C" at 1, f(1) center point pointtype 7 pointsize 2 offset -2
set label "I" at -1.25, f(-1.25) center point pointtype 7 pointsize 2 offset -2
plot (x < 0 ? f(x) : 1/0) ls 1 t "x^3 + 2/x", (x > 0 ? f(x) : 1/0) ls 1 t ""

\(g(x) = x + 4/x\)

\(g(x)\) is undefined at \(x = 0\)

Domain: \((-\infty, 0) \cup (0, \infty)\)

clear
pkg load symbolic
syms x
g = x + (4/x)

dg = simplify(diff(g,x))
dgg = simplify(diff(dg, x))

octave> octave> g = (sym)

      4
  x + ─
      x
octave> dg = (sym)

      4
  1 - ──
       2
      x
dgg = (sym)

  8
  ──
   3
  x

Both \(g'(x)\) and \(g''(x)\) are undefined at \(x = 0\). \(g'(x) = 0\) when \(x = -2, 2\). \(g''(x) = 0\) when \(x = \emptyset\)

function y = firstDerivative(x)
  num = 4
  den = x^2
  y = 1 - (num / den);
end

ans = [firstDerivative(-3), firstDerivative(-2), firstDerivative(-1), firstDerivative(1), firstDerivative(3)]

function y = secondDerivative(x)
  den = x^3
  y = 8 / den;
end

ans = [secondDerivative(-3), secondDerivative(-2), secondDerivative(-1), secondDerivative(1), secondDerivative(2), secondDerivative(3) ]

Let's check the function for vertical asymptotes.

\(\lim_{x \to 0^-} g(x) = -\infty\)

\(\lim_{x \to 0^+} g(x) = \infty\)

So \(x = 0\) is the vertical asymptote.

\(\lim_{x \to \pm \infty} g(x) = \pm \infty\)

So there is no horizontal asymptote.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
g(x) = x + (4/x)
set yrange [-20:20]
set xrange [-5:5]
set label "C" at -2, g(-2) left point pointtype 7 pointsize 2 offset -2
set label "C" at 2, g(2) left point pointtype 7 pointsize 2 offset -2
plot (x < 0 ? g(x) : 1/0) ls 1 t "x + 4/x", (x > 0 ? g(x) : 1/0) ls 1 t ""

\(f(x) = \dfrac{x^2 + x + 2}{x-1}\)

\(f(x)\) is undefined at \(x = 1\)

Domain: \((-\infty, 1) \cup (1, \infty)\)

clear
pkg load symbolic
syms x
f = (x^2 + x + 2)/(x-1)

df = simplify(diff(f,x))
dff = simplify(diff(df,x))

octave> octave> f = (sym)

   2
  x  + x + 2
  ──────────
    x - 1
octave> df = (sym)

   2
  x  - 2⋅x - 3
  ────────────
   2
  x  - 2⋅x + 1
dff = (sym)

     8⋅(x - 1)
  ───────────────
                2
  ⎛ 2          ⎞
  ⎝x  - 2⋅x + 1⎠

Both \(f'(x)\) and \(f''(x)\) are undefined at \(x=1\). \(f'(x) = 0\) when \(x = -1, 3\). \(f''(x) = 0\) when \(x = \emptyset\)

function y = firstDerivative(x)
  num = x^2 - (2*x) - 3
  den = (x-1)^2
  y = num / den;
end

ans = [firstDerivative(-2), firstDerivative(-1), firstDerivative(0), firstDerivative(2), firstDerivative(4)]

function y = secondDerivative(x)
  den = (x-1)^3;
  y = 8 / den;
end

ans = [secondDerivative(-2), secondDerivative(-1), secondDerivative(0), secondDerivative(2), secondDerivative(3), secondDerivative(4) ]

Let's check the function for vertical asymptotes:

\(\lim_{x \to 1^-} f(x) = -\infty\)

\(\lim_{x \to 1^+} f(x) = \infty\)

So \(x = 1\) is the vertical asymptote.

\(\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \dfrac{x(1 + 1/x + 2/x^2}{(1/x - 1/x^2} = \pm \infty\)

So there is no horizontal asymptote.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
f(x) = (x**2 + x + 2)/(x-1)
set yrange [-20:20]
set xrange [-2:4]
set label "C" at -1, f(-1) left point pointtype 7 pointsize 2 offset -2
set label "C" at 3, f(3) left point pointtype 7 pointsize 2 offset -2
plot (x < 1 ? f(x) : 1/0) ls 1 t "x + 4/x", (x > 1 ? f(x) : 1/0) ls 1 t ""

\(h(x) = \dfrac{\sqrt{x}}{x+3}\)

Domain of \(\sqrt{3}\) is \([0, \infty)\)

\(h(x)\) is not defined when \(x = -3\)

Domain of \(h(x)\) is \([0, \infty)\)

clear
pkg load symbolic
syms x
h = sqrt(x)/(x+3)

dh = simplify(diff(h,x))
dhh = simplify(diff(dh,x))

octave> octave> h = (sym)

    √x
  ─────
  x + 3
octave> dh = (sym)

      3 - x
  ─────────────
              2
  2⋅√x⋅(x + 3)
dhh = (sym)

           ⎛ 2          ⎞
         3⋅⎝x  - 6⋅x - 3⎠
  ──────────────────────────────
     3/2 ⎛ 3      2            ⎞
  4⋅x   ⋅⎝x  + 9⋅x  + 27⋅x + 27⎠

\(h'(x)\) is not defined when \(x = 0\). \(h''(x)\) is not defined when \(x = 0\). \(h'(x) = 0\) when \(x = 3\). \(h''(x) = 0\) when \(x = 3 + 2\sqrt{3}\)

function y = firstDerivative(x)
  num = 3 - x
  den = 2 * sqrt(x) * (x+3)^2
  y = num / den;
end

ans = [firstDerivative(2), firstDerivative(3), firstDerivative(3 + (2*sqrt(3))), firstDerivative(7)]

function y = secondDerivative(x)
  num = 3*(x * (x-6) - 3)
  den = 4*((x+3)^3)*(x^(1.5))
  y = num / den;
end

ans = [secondDerivative(2), secondDerivative(3), secondDerivative(7)]

Let's check the function for vertical asymptotes. There is no element \(x\) where \(h(x)\) is undefined in its domain. So it has no vertical asymptote.

\(\lim_{x \to \pm \infty} h(x) = \lim_{x \to \pm \infty} \dfrac{1}{\sqrt{x} + 3/\sqrt{x}} = 0\)

So \(y = 0\) is the horizontal asymptote.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "h(x)"
set grid
set key right top
h(x) = sqrt(x)/(x+3)
set xrange [-1:7]
set yrange [0:0.35]
set label "I" at 6.464101, h(6.464101) left point pointtype 7 pointsize 2 offset -2
set label "C" at 3, h(3) left point pointtype 7 pointsize 2 offset -2
plot (x > 0 ? h(x) : 1/0) ls 1 t "x^(1/2)/x+3", 0 ls 2 t "asymptote"

\(f(x) = \dfrac{x+1}{\sqrt{1-x^2}}\)

In order for \(f(x)\) to be defined, we must have \(1 - x^2 > 0\). To solve this inequality we first determine when \(1 - x^2 = 0\)

\(1 - x^2 = (1-x)(1+2) = 0\)

\(x = 1, -1\)

Now let's check the sample points in the intervals \((-\infty, -1), (-1,1), (1,\infty)\)

Thus the domain of \(f\) is \((-1,1)\)

clear
pkg load symbolic
syms x
f = (x+1)/sqrt(1-x^2)

df = simplify(diff(f,x))
dff = simplify(diff(df,x))

octave> octave> f = (sym)

     x + 1
  ───────────
     ________
    ╱      2
  ╲╱  1 - x
octave> df = (sym)

          -1
  ───────────────────
     ________
    ╱      2
  ╲╱  1 - x  ⋅(x - 1)
dff = (sym)

     2
  - x  - x⋅(x - 1) + 1
  ────────────────────
          3/2
  ⎛     2⎞           2
  ⎝1 - x ⎠   ⋅(x - 1)

Both \(f'(x)\) and \(f''(x)\) are undefined at \(x = 1, -1\)

function y = firstDerivative(x)
  num = 1
  den = (1-x)*sqrt(1-(x^2))
  y = num / den;
end

ans = [firstDerivative(0), firstDerivative(0.5), firstDerivative(-0.5), firstDerivative(-0.6)]

function y = secondDerivative(x)
  num = (2*x + 1)*(1+x)
  den = (1 - (x^2))^(5/2)
  y = num / den;
end

ans = [secondDerivative(0), secondDerivative(-0.3), secondDerivative(-0.6), secondDerivative(0.5)]

\(f'(x) = 0\) when \(x = \emptyset\)

\(f''(x) = 0\) when \(x = -1/2\)

Let's check the function for vertical asymptotes

$lim_{x → -1^+} f(x) = \(\lim_{x \to -1^+} \dfrac{x(1+1/x)}{x\sqrt{1/x^2 - 1}} = \dfrac{1-1}{-1\sqrt{-1 - 1}} = -\)

\(\lim_{x \to -1^-} f(x) = 0\)

\(\lim_{x \to 1^+} f(x) = \dfrac{1+1}{1(1-1)} = \infty\)

\(\lim_{x \to 1^-} f(x) = \infty\)

So \(x=1\) are vertical asymptotes

\(\lim_{x \to \infty} f(x) = \dfrac{1+1}{\sqrt{-2}} = -i\)

\(\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \sqrt{\dfrac{(1+x)^2}{(1-x^2)}} = \lim_{x \to -\infty} \sqrt{\dfrac{1+x}{1-x}}\)

\(= \lim_{x \to -\infty} \sqrt{\dfrac{0+1}{0-1}} = \sqrt{-1} = i\)

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
f(x) = (x+1)/sqrt(1-(x**2))
set xrange [-1:1]
set yrange [-1:6]
set label "I" at -0.5, f(-0.5) left point pointtype 7 pointsize 2 offset -2
plot f(x) ls 1 t "(x+1)/sqrt(1-x^2)"

\(g(x) = \dfrac{x+1}{\sqrt{x^2 - 1}}\)

In order for \(g(x)\) to be defined, we must have \(x^2 - 1 > 0\). To solve this inequality we first determine when \(x^2 - 1 = 0\)

\(x^2 - 1 = (x-1)(x+1) = 0\)

\(x = 1, -1\)

now we check sample points in the intervals \((-\infty, -1), (-1,1)\) and \((1, \infty)\)

Thus the domain of \(h\) is \((-\infty, -1) \cup (1, \infty)\)

Single differentiation:

pkg load symbolic
syms x
g = (x+1)/sqrt(x^2 - 1)
ans = simplify(diff(g,x))

octave> g = (sym)

     x + 1
  ───────────
     ________
    ╱  2
  ╲╱  x  - 1
ans = (sym)

          -1
  ───────────────────
             ________
            ╱  2
  (x - 1)⋅╲╱  x  - 1

Double differentiation:

pkg load symbolic
syms x
g = (x+1)/sqrt(x^2 - 1)
simplify(diff(g,x,x))

octave> g = (sym)

     x + 1
  ───────────
     ________
    ╱  2
  ╲╱  x  - 1
ans = (sym)

             2⋅x + 1
  ─────────────────────────────
     ________
    ╱  2      ⎛ 3    2        ⎞
  ╲╱  x  - 1 ⋅⎝x  - x  - x + 1⎠

Now let's find where \(g'(x)\) and \(g''(x)\) are \(0\).

pkg load symbolic
syms x
g = (x+1)/sqrt(x^2 - 1)
solve(diff(g,x) == 0, x)

octave> g = (sym)

     x + 1
  ───────────
     ________
    ╱  2
  ╲╱  x  - 1
ans = {}(0x0)

pkg load symbolic
syms x
g = (x+1)/sqrt(x^2 - 1)
solve(diff(g,x, x) == 0, x)

octave> g = (sym)

     x + 1
  ───────────
     ________
    ╱  2
  ╲╱  x  - 1
ans = (sym) -1/2

pkg load symbolic
syms x
g = (x+1)/sqrt(x^2 - 1)
dg = simplify(diff(g,x))
dgh = function_handle(dg)
[dgh(-2), dgh(-0.5), dgh(0), dgh(2)]

ddg = simplify(diff(g,x,x))
ddgh = function_handle(ddg)
[ddgh(-2), ddgh(-0.5), ddgh(0), ddgh(1), ddgh(2)]

octave> g = (sym)

     x + 1
  ───────────
     ________
    ╱  2
  ╲╱  x  - 1
dg = (sym)

          -1
  ───────────────────
             ________
            ╱  2
  (x - 1)⋅╲╱  x  - 1
dgh =

@(x) -1 ./ ((x - 1) .* sqrt (x .^ 2 - 1))
ans =

 Columns 1 through 3:

   0.19245 + 0.00000i  -0.00000 - 0.76980i  -0.00000 - 1.00000i

 Column 4:

  -0.57735 + 0.00000i
octave> ddg = (sym)

             2⋅x + 1
  ─────────────────────────────
     ________
    ╱  2      ⎛ 3    2        ⎞
  ╲╱  x  - 1 ⋅⎝x  - x  - x + 1⎠
ddgh =

@(x) (2 * x + 1) ./ (sqrt (x .^ 2 - 1) .* (x .^ 3 - x .^ 2 - x + 1))
warning: division by zero
warning: called from
    eval>@<anonymous> at line 1 column 15
ans =

 Columns 1 through 3:

   0.19245 + 0.00000i   0.00000 + 0.00000i   0.00000 - 1.00000i

 Columns 4 and 5:

       Inf + 0.00000i   0.96225 + 0.00000i

Let's check the function for vertical asymptotes

\(\lim_{x \to -1^-} g(x) = \lim_{x \to -1^-} \dfrac{x+1}{\sqrt{x^2 - 1}} = \lim_{x \to -1^-} \dfrac{\sqrt{x+1}}{\sqrt{x-1}} = 0\)

\(\lim_{x \to -1^+} g(x) = 0\)

\(\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} \dfrac{\sqrt{x+1}}{\sqrt{x-1}} = \lim_{x \to 1^-} \dfrac{x^2 - 1}{x-1} = -\infty\)

\(\lim_{x \to 1^+} g(x) = \infty\)

Since \(x = 1\) is a vertical asymptote.

\(\lim_{x \to \pm \infty} g(x) = \lim_{x \to \pm \infty} g(x) \sqrt{\dfrac{(x+1)^2}{x^2 -1}} = \lim_{x \to \pm \infty} \sqrt{\dfrac{x+1}{x-1}}\)

\(= \lim_{x \to \pm \infty} \sqrt{\dfrac{1 + 1/x}{1-1/x}} = \pm 1\)

So \(y = \pm 1\) are vertical asymptootes.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
g(x) = (x+1)/sqrt((x**2) - 1)
set xrange [-5:5]
# set yrange [-1:6]
plot (x < -1 ? g(x): 1/0) ls 1 t "(x+1)/sqrt(x^2 - 1)", (x > 1 ? g(x): 1/0) ls 1 t "", 1 ls 2 t "asymptote"

\(f(x) = \dfrac{(x-5)\sqrt{x}}{4}\)

Domain of \(f(x)\): \([0, \infty)\)

pkg load symbolic
syms x
f = ((x-5)*sqrt(x))/4
df = simplify(diff(f,x))
ddf = simplify(diff(f,x,x))
solve(diff(f,x) == 0, x)
solve(diff(f,x,x) == 0, x)

dfh = function_handle(df)
[dfh(1), dfh(5/3), dfh(2)]

ddfh = function_handle(ddf)
[ddfh(1), ddfh(5/3), ddfh(2)]

octave> f = (sym)

  √x⋅(x - 5)
  ──────────
      4
df = (sym)

  3⋅x - 5
  ───────
    8⋅√x
ddf = (sym)

  3⋅x + 5
  ───────
      3/2
  16⋅x
ans = (sym) 5/3
ans = (sym) -5/3
octave> dfh =

@(x) (3 * x - 5) ./ (8 * sqrt (x))
ans =

  -0.25000   0.00000   0.08839
octave> ddfh =

@(x) (3 * x + 5) ./ (16 * x .^ (3 / 2))
ans =

   0.50000   0.29047   0.24307

Let's check the function for vertical asymptotes

\(\lim_{x \to 0^+} f(x) = 0\)

\(\lim_{x \to \infty} f(x) = 0\)

\(\lim_{x \to -\infty} f(x) = -i\infty\)

So there are no horizontal or vertical asymptote.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
f(x) = (x-5)*sqrt(x)/4
set xrange [-0.5:9]
set label "C" at (1.667), f(1.667) left point pointtype 7 pointsize 2 offset -2
plot (x >= 0 ? f(x): 1/0) ls 1 t "(x-5)*sqrt(x)/4"

\(g(x) = \dfrac{(x-5)^2\sqrt{x}}{4}\)

Domain of \(g(x)\): \([0, \infty)\)

pkg load symbolic
syms x
g = (((x-5)^2) *sqrt(x))/4
dg = simplify(diff(g,x))
ddg = simplify(diff(g,x,x))

solve(diff(g,x) == 0, x)
solve(diff(g,x,x) ==0, x)

dgh = function_handle(dg)
[dgh(1), dgh(2), dgh(1+ ((2*sqrt(6))/3)), dgh(3), dgh(5), dgh(6)]

ddgh = function_handle(ddg)
[ddgh(1), ddgh(2), ddgh(3), ddgh(5), ddgh(6)]

octave> g = (sym)

            2
  √x⋅(x - 5)
  ───────────
       4
dg = (sym)

  5⋅(x - 5)⋅(x - 1)
  ─────────────────
         8⋅√x
ddg = (sym)

    ⎛   2          ⎞
  5⋅⎝3⋅x  - 6⋅x - 5⎠
  ──────────────────
           3/2
       16⋅x
octave> ans = (sym 2×1 matrix)

  ⎡1⎤
  ⎢ ⎥
  ⎣5⎦
ans = (sym 2×1 matrix)

  ⎡    2⋅√6⎤
  ⎢1 - ────⎥
  ⎢     3  ⎥
  ⎢        ⎥
  ⎢    2⋅√6⎥
  ⎢1 + ────⎥
  ⎣     3  ⎦
octave> dgh =

@(x) 5 * (x - 5) .* (x - 1) ./ (8 * sqrt (x))
ans =

  -0.00000  -1.32583  -1.48881  -1.44338   0.00000   1.27578
octave> ddgh =

@(x) 5 * (3 * x .^ 2 - 6 * x - 5) ./ (16 * x .^ (3 / 2))
ans =

  -2.50000  -0.55243   0.24056   1.11803   1.42462

\(g'(x) = -\) when \(x = 1,5\)

\(g''(x) = 0\) when \(x = 1- 2\sqrt{6}/3, 1+ 2\sqrt{6}/3\)

Let's check the function for vertical asymptote

\(\lim_{x \to 0} g(x) = 0\)

There is no vertical asymptote.

\(\lim_{x \to \infty} g(x) = \infty\)

\(\lim_{x \to -\infty} g(x) = i\infty\)

There is no horizontal asymptote.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"
reset session
set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
g(x) = ((x-5)**2)*sqrt(x)/4
set xrange [0:9]
set label "C" at 1, g(1) left point pointtype 7 pointsize 2 offset -2
set label "C" at 5, g(5) left point pointtype 7 pointsize 2 offset -2
set label "I" at 2.63, g(2.63) left point pointtype 7 pointsize 2 offset -2
plot g(x) ls 1 t "(x-5)^2*sqrt(x)/4"

Drawing graph for cubic root isn't straightforward.

\(f(x) = x - 3\sqrt[3]{x}\)

Domain of \(f(x)\) is \((-\infty, \infty)\)

pkg load symbolic
syms x
f = x - (3*nthroot(x,3))
df = simplify(diff(f,x))
ddf = simplify(diff(f,x,x))

solve(df == 0,x)
solve(ddf == 0, x)

dfh = function_handle(df)
[dfh(2)]

ddfh = function_handle(ddf)
[ddfh(1), ddfh(2)]

octave> f = (sym)

      3 ___
  - 3⋅╲╱ x  + x
df = (sym)

       1
  1 - ────
       2/3
      x
ddf = (sym)

    2
  ──────
     5/3
  3⋅x
octave> ans = (sym) 1
ans = {}(0x0)
octave> dfh =

@(x) 1 - 1 ./ x .^ (2 / 3)
ans =  0.37004
octave> ddfh =

@(x) 2 ./ (3 * x .^ (5 / 3))
ans =

   0.66667   0.20999

\(f'(x) = 0\) when \(x=1\)

\(f''(x) = 0\) when \(x=\emptyset\)

The function \(f(x)\) doesn't have vertical asymptotes.

\(\lim_{x \to \pm \infty} f(x) = \pm \infty\)

There are no horizontal asymptotes.

Drawing graph for cubic root isn't straightforward via gnuplot.

\(f(x) = \sqrt[3]{x}(\sqrt[3]{x} - 2)\)

Domain: \((-\infty, \infty)\)

pkg load symbolic
syms x
f = nthroot(x,3)*(nthroot(x,3)-2)
df = simplify(diff(f,x))
ddf = simplify(diff(f,x,x))

solve(df == 0,x)
solve(ddf == 0, x)

dfh = function_handle(df)
[dfh(1),dfh(7),dfh(8),dfh(9)]

ddfh = function_handle(ddf)
[ddfh(1), ddfh(7), ddfh(8), ddfh(9)]

octave> f = (sym)

  3 ___ ⎛3 ___    ⎞
  ╲╱ x ⋅⎝╲╱ x  - 2⎠
df = (sym)

    ⎛3 ___    ⎞
  2⋅⎝╲╱ x  - 1⎠
  ─────────────
         2/3
      3⋅x
ddf = (sym)

    ⎛    3 ___⎞
  2⋅⎝2 - ╲╱ x ⎠
  ─────────────
         5/3
      9⋅x
octave> ans = (sym) 1
ans = (sym) 8
octave> dfh =

@(x) 2 * (x .^ (1 / 3) - 1) ./ (3 * x .^ (2 / 3))
ans =

   0.00000   0.16632   0.16667   0.16642
octave> ddfh =

@(x) 2 * (2 - x .^ (1 / 3)) ./ (9 * x .^ (5 / 3))
ans =

   0.22222   0.00076   0.00000  -0.00046

The function doesn't have any vertical asymptotes.

\(\lim_{x \to \pm \infty} f(x) = \infty\)

The function doesn't have horizontal asymptotes.

Drawing graph for cubic root isn't straightforward via gnuplot.

\(g(x) = 3x^(2/3) - 2x\)

Domain: \((-\infty, \infty)\)

pkg load symbolic
syms x
f = 3*cbrt(x^2) - (2*x)
df = simplify(diff(f,x))
ddf = simplify(diff(f,x,x))

solve(df == 0,x)
solve(ddf == 0, x)

dfh = function_handle(df)
[dfh(-1), dfh(0), dfh(2)]

ddfh = function_handle(ddf)
[ddfh(-1), ddfh(0), ddfh(1), ddfh(2)]

octave> f = (sym)

              ____
           3 ╱  2
  -2⋅x + 3⋅╲╱  x
df = (sym)

            ____
         3 ╱  2
       2⋅╲╱  x
  -2 + ─────────
           x
ddf = (sym)

        ____
     3 ╱  2
  -2⋅╲╱  x
  ───────────
         2
      3⋅x
octave> ans = (sym) 1
ans = {}(0x0)
octave> dfh =

@(x) -2 + 2 * (x .^ 2) .^ (1 / 3) ./ x
warning: division by zero
warning: called from
    eval>@<anonymous> at line 1 column 9
ans =

  -4.00000       NaN  -0.41260
octave> ddfh =

@(x) -2 * (x .^ 2) .^ (1 / 3) ./ (3 * x .^ 2)
warning: division by zero
warning: called from
    eval>@<anonymous> at line 1 column 22
ans =

  -0.66667       NaN  -0.66667  -0.26457

Suppose that \(f\) is a function whose domain contains the interval \((-\infty, b)\) and let \(A= { f(x): x < b}\).

Theorem: If \(f\) is weakly decreasing on \((-\infty, b)\) and \(A\) has an upper bound, then \(\lim_{x -> -\infty}\) is defined and is equal to the least upper bound of \(A\).

Proof: Suppose \(f\) is weakly decreasing and \(A\) has an upper bound. By the completeness of the real number, we can let \(L\) be the least upper bound of \(A\). We now verify that \(\lim_{x \to -\infty} f(x) = L\) by using the definition of limits.

Suppose \(\epsilon > 0\). Since \(L\) is the least upper bound of \(A\), \(L - \epsilon\) is not an upper bound and therefore there must be some number \(N < b\) such that \(f(N) > L-\epsilon\). Now consider any \(x < N\). Since \(f\) is weakly decresing on \((-\infty, b)\), \(f(x) \geq f(N) > L - \epsilon\), and since \(L\) is an upper bound for \(A\), \(f(x) \leq L\). Thus, \(L - \epsilon < f(n) \leq f(x) \leq L < L + \epsilon\), so \(|f(x)-L| < \epsilon\).