set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot [0:5] x**2 - (6*x) + 5 title "x^2 - 6x + 5" ls 1

\(f(x) = x^2 - 6x + 5\)

\(f'(x) = 2x-6\)

There are no points where \(f\) is not differentiable but there is one point where derivative is 0.

\(x = 3\)

By the first derivative test, \(f\) has a local minimum at \(3\). Function \(f\) has no local maximum.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot [0:5] 3*x - x**2 title "3x - x^2" ls 1

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

\(g'(x) = 3 - 2x\)

There are no points where \(g\) is not differentiable but there is one point where the derivative is \(0\).

\(x = 1.5\)

By the first derivative test, \(f\) has a local maximum at \(1.5\). Function \(g\) has no local minimum.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot 12*x - x**3 title "12x - x^3" ls 1

Viewing local minimum:

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot [-4:0] 12*x - x**3 title "12x - x^3" ls 1

Viewing local maximum:

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot [-2:3] 12*x - x**3 title "12x - x^3" ls 1

\(f(x) = 12x - x^3\)

\(f'(x) = 12 - 3x^2\)

There are no points where \(f\) is not differentiable but there is two point where the derivative is zero.

\(x = 2, -2\)

By the first derivative test, \(f\) has a local maximum at \(2\) and local minimum at \(-2\).

Graph of the function:

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
plot (x**2 - 1)**2 title "(x^2 - 1)^2" ls 1

Graph for understanding their local minimum/maximum better:

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
plot [-2:2] (x**2 - 1)**2 title "(x^2 - 1)^2" ls 1

\(g(x) = (x^2 - 1)^2\)

\(g(x) = x^4 + 1 - 2x^2\)

\(g'(x) = 4x^3 - 4x\)

\(= 4x(x^2 - 1)\)

There are no points where \(g\) is not differentiable but there are three points where the derivate is zero.

\(x = 0, 1, -1\)

By the first derivative test, the local maximum is attained at \(0\) and local minimum is attained at \(-1\) and \(1\).

Graph of the function:

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot ((x**4 / 2) - 4*x**3 + 9*x**2 - 5) title "x^4/2 - 4x^3 + 9x^2 - 5" ls 1

\(f(x) = \dfrac{x^4}{2} - 4x^3 + 9x^2 - 5\)

\(f'(x) = 2x^3 - 12x^2 + 18x\)

There are no points where \(f\) is not differentiable but there are points where the derivate is zero.

\(f'(x) = x(2x^2 - 12x + 18)\)

\(f'(x) = 2x(x^2 - 6x + 9)\)

\(x = 0,3\)

By the first derivative test, the local minimum is attained at \(0\) and there is no local maximum.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
plot ((x**4 / 2) - 4*x**3 + 5*x**2 - 5) title "x^4/2 - 4x^3 + 5x^2 - 5" ls 1

\(g(x) = \dfrac{x^4}{2} - 4x^3 + 5x^2 - 5\)

\(g'(x) = 2x^3 - 12x^2 + 10x\)

\(= 2x(x^2 - 6x + 5)\)

There are no points where \(f\) is not differentiable but there are three points where derivative is zero.

\(x = 0,1,5\)

By the first derivative test, the local minimum is attained at \(5,0\) and local maximum is attained at \(1\).

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot x - 4* sqrt(x) title "x - 4*sqrt(x)" ls 1

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

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

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

There is one point where \(f\) is not differentiable.

\(x = 0\)

There is on point where the deriviate is zero. \(x = 4\).

By the first derivative test we have local maximum at \(4\).

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot x**2 - 4* sqrt(x) title "x^2 - 4*sqrt(x)" ls 1

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

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

\(g'(x) = 2x - \dfrac{2}{\sqrt{x}}\)

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

There is one point where \(g\) is not differentiable. \(x = 0\)

There is one point where the derivative is zero. \(x = 1\)

By the first derivative test, the local minimum is attained at \(1\).

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "f(x)"
set grid
set key right top
plot 1 / (x**2 + 1) title "1 / (x^2 + 1)" ls 1

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

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

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

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

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

There is no point where \(f\) is not differentiable. There is one point where the derivative is zero. \(x = 0\)

By first derivative test, the local maximum is attained at zero.

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "g(x)"
set grid
set key right top
plot x / (x**2 + 1) title "x / (x^2 + 1)" ls 1

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

\(g'(x) = \dfrac{x.-1.2x}{(x^2 + 1)^2} + (x^2 + 1)^{-1}\)

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

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

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

There is no point where \(g\) is not differentiable. There is two points where the derivative is zero. \(x = 1, -1\)

By first derivative test, the local minimum is attained at \(-1\) and the local maximum is attained at \(1\).

set terminal png notransparent nointerlace rounded font "Alegreya, 14"

set xlabel "x"
set ylabel "h(x)"
set grid
set key right top
plot x**2 / (x**2 + 1) title "x^2 / (x^2 + 1)" ls 1

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

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

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

\(h'(x) = x^2.-1.(x^2 + 1)^{-2}.2x + (x^2 + 1)^{-1}.2x\)

\(= \dfrac{-2x^3}{(x^2 + 1)^2} + \dfrac{2x}{(1+x^2)}\)

\(= \dfrac{-2x^3 + 2x(1+x^2)}{(1+x^2)^2}\)

\(= \dfrac{2x + 2x^3 - 2x^3}{(1+x^2)^2}\)

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

There is no point where \(h\) is not differentiable. There is one point where the derivative is zero. \(x=0\)

By first derivative test, the local minimum is attained at \(0\). It has no local maximum.

\(a_n = (\dfrac{n+1}{n^2+3})_{n=1}^{\infty}\)

Let's study the function

\(f(x) = \dfrac{x+1}{x^2 + 3}\)

Domain: \([1, \infty)\)

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

\(f'(x) = (x+1).-1.(x^2+3)^{-2}.2x + (x^2 + 3)^{-1}.1\)

\(= \dfrac{-1(x+1)2x}{(x^2+3)^2} + \dfrac{1}{x^2 + 3}\)

\(= \dfrac{-2x(x+1)}{(x^2+3)^2} + \dfrac{x^2 + 3}{(x^2 + 3)^2}\)

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

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

\(\forall x \geq 1; (x^2 + 3)^2 > 0\)

\(3 - x^2 - 2x \leq 0\)

\(\dfrac{3-x^2-2x}{(x^2+3)^2} \leq 0\)

Thus \(f\) is weakly decreasing in the interval \([1, \infty)\). This means that for any real number \(x_1\) and \(x_2\) with \(1 \leq x_1 < x_2\), \(f(x_1) > f(x_2)\). In particular, for every positive integer \(f(n) > f(n+1)\). But \(f(n) = a_n\), so this means for every positive integer \(a_n > a_{n+1}\). In other words, \((a)_{n=1}^{\infty}\) is strictly decreasing.

\(\left( \dfrac{n-1}{n^2+8} \right)^{\infty}_{n=1}\)

Let's study the function

\(f(x) = \dfrac{n-1}{n^2+8}\)

Domain: \([1, \infty)\)

\(f(x) = (x-1)(x^2+8)^{-1}\)

\(f'(x) = (x-1).-1.(x^2+8)^{-2}.2x + (x^2+8)^{-1}(1)\)

\(= \dfrac{-(x-1).2x}{(x^2+8)^2} + \dfrac{1}{x^2 + 8}\)

\(= \dfrac{-2x(x-1)}{(x^2+8)^2} + \dfrac{x^2+8}{(x^2+8)^2}\)

\(= \dfrac{x^2+8-2x^2+2x}{(x^2+8)^2}\)

\(= \dfrac{-x^2+2x+8}{(x^2+8)^2}\)

\(= \dfrac{2x-x^2+8}{(x^2+8)^2}\)

\(\forall x \geq 1\)

\((x^2 + 8)^2 > 0\)

\((2x - x^2 + 8) < 10\)

So, \(a > 0 \implies \dfrac{1}{1} > 0\)

\(a > 0 \land b > 0 \implies ab > 0\)

\(a > 0 \land b < 0 \implies ab < 0\)

So for \([1,4)\)

\(2x-x^2 + 8 > 0\)

And \([4, \infty)\)

\(2x-x^2 + 8 < 0\)

So the function is both increasing and decreasing with the domain \([1, \infty)\). So the sequence is not monotone.

\(\left( \dfrac{1}{n^2} - \dfrac{1}{n} \right)^{\infty}_{n=1}\)

Let's study the function

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

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

\(f'(x) = (1-x).-2.x^{-3} + x^{-2}.-1\)

\(= (x-1)2x^{-3} - \dfrac{1}{x^2}\)

\(= \dfrac{2(x-1)}{x^3} - \dfrac{1}{x^2}\)

\(= \dfrac{2x-2-x}{x^3} = \dfrac{x-2}{x^3}\)

Domain: \([1, \infty)\)

\(\forall x \geq 1; x^3 > 0\)

For \(x \in [1,2); x-2 < 0\)

\(x \in (2,\infty); x-2 > 0\)

\(x = 2, 0\)

So the function is both increasing and decreasing with the domain \([1,\infty)\). So the sequence is not monotone.

\(\left( \dfrac{1}{n^2} - \dfrac{4}{n} \right)^{\infty}_{n=1}\)

Let's study the function

\(f(x) = \dfrac{1}{x^2} - \dfrac{4}{x}\)

Domain: \([1, \infty)\)

\(f(x) = \dfrac{1}{x^2} - \dfrac{4x}{x^2} = \dfrac{1-4x}{x^2} = (1-4x)x^{-2}\)

\(f'(x) = (1-4x).-2x^{-3}+x^{-2}.-4\)

\(= \dfrac{8x-2}{x^3} - \dfrac{4}{x^2}\)

\(= \dfrac{8x-2-4x}{x^3}\)

\(= \dfrac{4x-2}{x^3}\)

\(\forall x \geq 1; x^3 > 0\)

\(\forall x \geq 1; 4x-2 > 0\)

So, \(\dfrac{4x-2}{x^3} > 0\)

Therefore \(f\) is strictly increasing on the interval \([1, \infty)\). This means that for any real number \(x_1\) and \(x_2\) with \(1 \leq x_1 \leq x_2\), \(f(x_1) < f(x_2)\). In particular, for every positive integer \(n\), \(f(n) < f(n+1)\). But \(f(n) = a_n\), so that means that for every positive integer \(n\), \(a_n < a_{n+1}\). In other words, \((a_n)^{\infty}_{n=1}\) is strictly increasing.

\(\left( \dfrac{2n^3}{3} - 5n^2 + 12n \right)_{n=1}^{\infty}\)

Let's study the function

\(f(x) = \dfrac{2x^3}{3} - 5x^2 + 12x\)

Domain: \([1, \infty)\)

\(f'(x) = 2x^2 - 10x + 12\)

\(\forall x \geq 1 2x^2 > 0\)

\(\forall x \geq 1 12 > 0\)

\(\forall x \geq 1 -10x < 0\)

For \(x \in [1,5)\), \(x-5 < 0\)

\(x=5\); \(x-5 = 0\)

\(x \in (5, \infty); x-5 > 0\)

So the function is both increasing and decreasing within the domain \([1, \infty)\). So the sequence is not monotone.

\(\left( \dfrac{4n^3}{3} - 9n^2 + 20n \right)_{n=1}^{\infty}\)

Let's study the function

\(f(x) = \dfrac{4x^3}{3} - 9x^2 + 20x\)

\(f'(x) = 4x^2 - 18x + 20\)

Domain: \([1, \infty)\)

\(f'(x) = 2x(2x-9) + 20\)

\(x \in [1,4.5); 2x-9 < 0\)

\(x=4.5; 2x-9=0\)

\(x \in (4.5, \infty); 2x-9 > 0\)

So the function is both increasing and decreasing within the domain \([1, \infty)\). So the sequence is not monotone.

\(\left( \sqrt{n^2 + 2} -n \right)_{n=1}^{\infty}\)

Let's study the function

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

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

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

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

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

Domain: \([1, \infty)\)

\(\forall x \geq 1; x > 0\)

\(\forall x \geq 1; \sqrt{x^2 + 2} > 0\)

So, \(\forall x \geq 1\)

\(\dfrac{x}{\sqrt{x^2 + 2}} > 0\)

When \(x=1\), \(\dfrac{x}{\sqrt{x^2 + 2}} = \dfrac{1}{\sqrt{3}} \approx 0.57\)

\(0.57 - 1 < 0\)

When \(x=2\), \(\dfrac{x}{\sqrt{x^2 + 2}} = \dfrac{2}{\sqrt{6}} \approx 0.81\)

\(0.81 - 1 < 0\)

We know that

\(x^2 + 2 > x^2\)

\(\sqrt{x^2 + 2} > x\)

\(1 > \dfrac{x}{x^2 + 2}\)

\(0 > \dfrac{x}{x^2 + 2} - 1\)

So, \(\dfrac{x}{\sqrt{x^2 + 2}} - 1 < 0\)

Thus \(f\) is strictly decreasing on the interval \([1, \infty)\). This measns that for any real numbers \(x_1\) and \(x_2\) with \(1 \leq x_1 \leq x_2\), \(f(x_1) > f(x_2)\). In particular, for every positive integer \(n\), \(f(n) > f(n+1)\). But \(f(n) = a_n\), so this means for every positive integer \(n\), \(a_n > a_{n+1}\). In other words, \((a_n)^{\infty}_{n=1}\) is strictly decreasing.

\(\left( \sqrt{n^2 + 2n} - n \right)^{\infty}_{n=1}\)

Let's study the function

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

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

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

\(x^2 + 1^2 + 2x > x^2 + 2x\)

\(x + 1 > \sqrt{x^2 + 2x}\)

\(\dfrac{x+1}{\sqrt{x^2 + 2x}} > 1\)

\(\dfrac{x+1}{\sqrt{x^2 + 2x}} - 1 > 0\)

Thus \(f\) is strictly increasing on the interval \([1, \infty)\). This means that for any real number \(x_1\) and \(x_2\) with \(1 \leq x_1 \leq x_2\), \(f(x_1) < f(x_2)\). In particular, for every positive integer \(n\), \(f(n) < f(n+1)\). But \(f(n) = a_n\), so that means that for every positive integer \(n\), \(a_n < a_{n+1}\). In other words, \((a_n)^{\infty}_{n=1}\) is strictly increasing.

\(a\) and \(b\) are real numbers.

\(f(x) = ax^4 + bx^3\)

Maximum value of \(f\) is 1.

\(f\) attains this maximum value at \(-1\).

\(f(-1) = a(-1)^4 + b(-1)^3\)

\(1 = a - b\)

\(f'(x) = 4ax^3 + 3bx^2\)

From theorem 4.3.6, we know that either \(f\) is not differentiable at \(-1\) or \(f'(-1) = 0\). We know that \(f'(x)\) is differentiable in it's domain.

\(f'(-1) = 4a(-1^3) + 3b(-1)^2\)

\(0 = -4a + 3b\)

\(0 = -a + 3b - 3a\)

\(0 = -a + 3(b-a)\)

\(0 = -a + 3.-1\)

\(a = -3\)

\(a-b=1\)

\(-3-b=1\)

\(b = -3-1 = -4\)

So, \(a=-3\) and \(b=-4\)

Domain of \(f\) is interval \(I\).

Domain of \(g\) is interval \(J\).

\(\forall x \in J g(x) \in I\)

Suppose that \(f'(x) \leq 0\) for all \(x\) in the interior of \(I\). To prove that \(f\) is weakly decreasing on \(I\), we assume that \(x_1\) and \(x_2\) are elements of \(I\) and \(x_1 < x_2\). According to definition 4.3.1, we must prove that \(f(x_1) \geq f(x_2)\)

We need to compare \(f(x_1)\) and \(f(x_2)\), so let's try applying the mean value theorem on the interval \([x_1, x_2]\). Notice that \([x_1, x_2] \subseteq I\). So since \(f\) is continous on \(I\) and differentiable on the interior of \(I\), it is continous on \([x_1, x_2]\) and differentiable on \((x_1, x_2)\). Therefore according to mean value thorem, there is some number \(c\) such that \(x_1 < c < x_2\) and

\(\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(c)\)

But we have assumed that \(f'(x) \leq 0\) for all \(x\) in the interior of \(I\). So \(f'(c) \leq 0\). Since \(x_1 < x_2\), it follows that \(f(x_2) - f(x_1) = f'(c)(x_2-x_1) \leq 0\)

and therefore \(f(x_2) \leq f(x_1)\)

Suppose that \(f'(x) = 0\) for all \(x\) in the interior of \(I\). We must prove \(f\) is constant on \(I\).

We assume that \(x_1\) and \(x_2\) are elements of \(I\). Let's try applying the mean value theorem to \(f\) on the interval \([x_1,x_2]\). Notice that \([x_1, x_2] \subseteq I\), so since \(f\) is continous on \(I\) and differentiable on the interior of \(I\), it is continous on \([x_1,x_2]\) and differentiable on \((x_1, x_2)\). Therefore according to mean value theorem, there is some number \(c\) such that \(x_1 < c < x_2\)

\(\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(c)\)

\(f(x_2) - f(x_1) = f'(c)(x_2 - x_1)\)

We have assumed that \(f'(c) = 0\) for all in the interior of \(I\), so

\(f(x_2) - f(x_1) = 0\)

\(f(x_2) = f(x_1)\)

So \(f\) is constant on \(I\).

Notice that in the proof we assumed that \(x_1 < x_2\). In case if it is \(x_1 > x_2\), it's the same proof with intervals reversed. In case it's \(x_1 = x_2\), then from the definition of the function it follows that \(f(x_1) = f(x_2)\).