\(f(x) = 2x^3 - 9x^2 + 3x - 1\)

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

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
f = 2*x^3 - 9*x^2 + (3*x) - 1

octave> octave> f = (sym)

     3      2
  2⋅x  - 9⋅x  + 3⋅x - 1

int(f, x)

ans = (sym)

   4             2
  x       3   3⋅x
  ── - 3⋅x  + ──── - x
  2            2

So the antiderivative is \(\dfrac{x^4}{2} - 3*x^3 + \dfrac{3x^2}{2} - 2\) on the interval \((-\infty, \infty)\).

\(g(x) = \sqrt{x} + \sqrt[3]{x}\)

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

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
g = sqrt(x) + (x)^(1/3)

octave> octave> warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 379 column 13
    mpower at line 76 column 5
g = (sym)

  3 ___
  ╲╱ x  + √x

int(g,x)

ans = (sym)

     4/3      3/2
  3⋅x      2⋅x
  ────── + ──────
    4        3

Intervals where the antiderivaties apply is same as the original function's domain.

\(f(x) = \sin x + \cos x\)

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

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
f = sin(x) + cos(x)
int(f,x)

octave> octave> f = (sym) sin(x) + cos(x)
ans = (sym) sin(x) - cos(x)

Intervals where the antiderivaties apply is same as the original function's domain.

\(g(x) = \sec x(\sec x + \tan x)\)

Let's find the domain of the above function.

We know than \(\sec x = \dfrac{1}{\cos x}\) and \(\tan x = \dfrac{\sin x}{\cos x}\)

So, \(g(x) = \sec^2 x + \sin x \sec^2 x\)

We know the domain of \(\sin x\) is \((-\infty, \infty)\). But domain of \(\sec x\) is \(R - \dfrac{\pi}{2} - n*\pi\)

So the domain of the function \(g(x)\) is \(R - \dfrac{\pi}{2} - n*\pi\).

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
g = sec(x)*(sec(x) + tan(x))
int(g,x)

octave> octave> g = (sym) (tan(x) + sec(x))⋅sec(x)
ans = (sym) tan(x) + sec(x)

Intervals where the antiderivaties apply is same as the original function's domain.

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

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

Let's solve this using GNU Octave:

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

octave> octave> f = (sym)

   2   1
  x  + ──
        2
       x
ans = (sym)

   3
  x    1
  ── - ─
  3    x

Intervals where the antiderivaties apply is same as the original function's domain.

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

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

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
g = (x^3 + 3)/(x^2)
int(g,x)

octave> octave> g = (sym)

   3
  x  + 3
  ──────
     2
    x
ans = (sym)

   2
  x    3
  ── - ─
  2    x

Intervals where the antiderivaties apply is same as the original function's domain.

\(f(x) = (x^3 + 3)^2\)

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

Let's solve this using GNU Octave:

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

octave> octave> f = (sym)

          2
  ⎛ 3    ⎞
  ⎝x  + 3⎠
ans = (sym)

   7      4
  x    3⋅x
  ── + ──── + 9⋅x
  7     2

Intervals where the antiderivaties apply is same as the original function's domain.

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

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

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
g = ((x^3 +3)/x)^2
int(g,x)

octave> octave> g = (sym)

          2
  ⎛ 3    ⎞
  ⎝x  + 3⎠
  ─────────
       2
      x
ans = (sym)

   5
  x       2   9
  ── + 3⋅x  - ─
  5           x

Intervals where the antiderivaties apply is same as the original function's domain.

\(f(x) = sqrt(2x+3)\)

Let's find it's domain:

\(2x + 3 \geq 0\)

\(x \geq \dfrac{-3}{2}\)

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
f = sqrt(2*x + 3)
int(f,x)

octave> octave> f = (sym)

    _________
  ╲╱ 2⋅x + 3
ans = (sym)

           3/2
  (2⋅x + 3)
  ────────────
       3

Intervals where the antiderivaties apply is same as the original function's domain.

\(f'(x) = 4x - 3\)

\(f(1) = 4\)

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
fd(x) = 4*x - 3
int(fd, x)

octave> octave> fd(x) = (symfun) 4⋅x - 3
ans(x) = (symfun)

     2
  2⋅x  - 3⋅x

So the antiderivate is \(2x^2 - 3x + C = F(x)\)

Now let's find the function,

syms c
f = 2*x^2 - 3*x + c
f_h = function_handle(f)
solve(f_h(c, 1) == 4, c)

f = (sym)

         2
  c + 2⋅x  - 3⋅x
f_h =

@(c, x) c + 2 * x .^ 2 - 3 * x
ans = (sym) 5

So the function is \(2x^2 - 3x + 5\)

\(g'(x) = \sqrt(x) + \sin x\)

\(g(0) = 1\)

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
gd(x) = sqrt(x) + sin(x)
int(gd, x)

octave> octave> gd(x) = (symfun) √x + sin(x)
ans(x) = (symfun)

     3/2
  2⋅x
  ────── - cos(x)
    3

So the antiderivative is \(\dfrac{2*x^{3/2}}{3} - \cos x + c = F(x)\)

Now let's find this function,

syms c
f = 2*(x)^(3/2)/3 - cos(x) + c
f_h = function_handle(f)
solve(f_h(c, 0) == 1, c)

warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 379 column 13
    mpower at line 76 column 5
f = (sym)

         3/2
      2⋅x
  c + ────── - cos(x)
        3
f_h =

@(c, x) c + 2 * x .^ (3 / 2) / 3 - cos (x)
ans = (sym) 2

So the function is \(\dfrac{2*x^{3/2}}{3} - \cos x + 2 = f(x)\)

\(f'(x) = \dfrac{1}{\sqrt{x}} + \sec x \tan x\)

\(\lim_{x \to 0^+} f(x) = 2\)

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
fd(x) = (1/sqrt(x) + sec(x)*tan(x))
int(fd, x)

octave> octave> fd(x) = (symfun)

                  1
  tan(x)⋅sec(x) + ──
                  √x
ans(x) = (symfun)

           1
  2⋅√x + ──────
         cos(x)

So the antiderivative is \(2\sqrt{x} + \dfrac{1}{\cos x} + c = F(x)\)

Now let's find this function,

syms c
f = 2*(x)^(1/2) + (1/cos(x)) + c
f_h = function_handle(f)
solve(f_h(c,0) == 2)

warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 379 column 13
    mpower at line 76 column 5
f = (sym)

               1
  c + 2⋅√x + ──────
             cos(x)
f_h =

@(c, x) c + 2 * sqrt (x) + 1 ./ cos (x)
ans = (sym) 1

So the function is \(2\sqrt{x} + \dfrac{1}{\cos x} + 1 = f(x)\)

\(g'(x) = x^2 \sqrt[3]{x}\)

\(g(1) = 2\)

Let's solve this using GNU Octave:

clear
pkg load symbolic
syms x
gd(x) = x^2 * x^(1/3)
int(gd, x)

octave> octave> warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 379 column 13
    mpower at line 76 column 5
gd(x) = (symfun)

   7/3
  x
ans(x) = (symfun)

     10/3
  3⋅x
  ───────
     10

So the antiderivative is \(\dfrac{3*x^(10/3)}{10} + c = G(x)\)

Now let's find this function,

syms c
g = 3*(x^(10/3))/10 + c
g_h = function_handle(g)
solve(g_h(c, 1) == 2)

warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 379 column 13
    mpower at line 76 column 5
g = (sym)

         10/3
      3⋅x
  c + ───────
         10
g_h =

@(c, x) c + 3 * x .^ (10 / 3) / 10
warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 379 column 13
    plus at line 61 column 5
    eval>@<anonymous> at line 1 column 10
ans = (sym)

  17
  ──
  10

So the function is \(\dfrac{3*x^(10/3)}{10} + \dfrac{17}{10} = G(x)\)

At \(t=0\), we are starting to apply brake. It's also \(h - 0\) and \(a = -10ft/s^2\)

\(h\) is the distance car travelled.

At \(t=t_1\), the car comes to stop.

\(h = 125\) feet

\(f''(t) = a = -10\)

\(f'(t) = -10t + c\)

We need to find \(f'(t)\) at \(t = 0\)

\(h = f(t) = -5*t^2 + c*t + d\)

At \(t=0, h=0\)

\(h = f(0) = d\)

So, \(d = 0\)

At \(t=t_1\), \(h = 125\)

\(h = f(t_1) = -5t_1^2 + ct_1\)

\(125 = -5t_1^2 + ct_1\)

Velocity at \(t_1 = 0\)

\(f'(t_1) = -10t_1 + c = 0\)

\(c = 10*t_1\)

So, \(125 = -5*t_1^2 + 10*t_1^2\)

\(5*t_1^2 = 125\)

\(t_1^2 = 25\)

\(t_1 = 5\)

So velocity at \(t=0\):

\(f'(0) = c = 10t_1 = 10.5 = 50\) feet/s

At \(t=0\), he applies brakes.

At \(t=2\), he travels 100 feets.

At t=0,

\(f''(t) = a = -10\) feet/s²

\(f'(t) = -10t + c\)

\(f(t) = -5t^2 + ct + d\)

We know that \(f(2) = 100\)

We need to find \(f'(t)\) at \(t = 0\)

We also know that \(f(0) = 0\)

So, \(f(0) = d = 0\)

So, \(d = 0\)

\(f(2) = 100\)

\(-5.2^2 + c.2 + 0 = 100\)

\(-20 + 2c = 100\)

\(2c - 120\)

\(c - 60\)

\(f'(t) = -10t + c\)

\(f'(0) = c = 60\)

So he was going at 60 feet/s when he applied the brakes.

Initial velocity = 10m/s

Broom acceleration = \(4 m/sec^2\)

Snitch distance = 100 m

Wwe need to find the time to reach snitch

\(a = f''(t) = 4\)

\(f'(t) = 4t + c\)

We know that \(f'(0) = 10\)

\(c = 10\)

\(f'(t) = 4t + 10\)

\(f(t) = 2t^2 + 10t + d\)

We need to find some \(t\) such that \(f(t) = 100\)

We know that at \(t = 0\), \(f(0) = 0\) and \(t = t\), \(f(t) = 100\)

From \(f(0) = 0\), we can conclude that \(d = 0\)

\(f(t) = 100\)

\(2t^2 + 10t = 100\)

\(t^2 + 5t = 50\)

\(t^2 + 5t - 50 - 0\)

So at \(t = 5\) seconds, harry will reach the snitch.

Tank capacity \(= 500\) Gallon

Water started leaking at noon

At \(t\) minutes after noon, water started leaking at \(5 + 2t\) gallons per miniutes.

\(f(t) = 5 + 2t\)

We need to find at what \(t\), \(f(t)\) becomes \(500\).

\(5 + 2t = 500\)

\(2t = 495\)

\(t = 247.5\)

At \(247.5\) minutes after noon, the tank becomes empty.