Let us find the derivative of the given function with respect to $x$:
$$f'(x) = \frac{d}{dx}(x^3 - x) = 3x^2 - 1$$
We know that the slope of the tangent line is equal to this derivative. Let us substitute the coordinate $x = 2$ into our expression:
$$\text{Slope } m = f'(2) = 3(2)^2 - 1 = 3(4) - 1 = 11$$

Let us calculate the first derivative of our function with respect to $x$:
$$f'(x) = \frac{d}{dx}(7x - 3) = 7$$
Now, we analyze the sign of the derivative. Since $7$ is a positive constant, we have $f'(x) > 0$ for all real values of $x$.

Therefore, the function is strictly increasing.

Let us write down the derivative of the function:
$$f'(x) = \frac{d}{dx}(x^2 - 6x + 8) = 2x - 6$$
To find the critical points, we set this derivative equal to zero:
$$2x - 6 = 0 \implies 2x = 6 \implies x = 3$$
Therefore, the critical point is at $x = 3$.

Let us compute the first derivative of the given function:
$$f'(x) = \frac{d}{dx}(-x^2 + 4x) = -2x + 4$$
For the function to be decreasing, we must have the slope be negative ($f'(x) < 0$):
$$-2x + 4 < 0 \implies -2x < -4$$
Dividing both sides by $-2$ and reversing the inequality sign gives:
$$x > 2$$
Therefore, the function is decreasing on the interval $(2, \infty)$.

Let us state the criteria for a local minimum of a differentiable function $f(x)$ at a point $x = c$.

1. The first derivative must be zero, indicating a stationary flat slope:
$$f'(c) = 0$$
2. The second derivative must be strictly positive, indicating that the curve is concave upward:
$$f''(c) > 0$$

Let us differentiate the function successively. First, let us apply the Power Rule to find the first derivative:
$$f'(x) = \frac{d}{dx}(x^4) = 4x^3$$
Now, let us differentiate again to find the second derivative:
$$f''(x) = \frac{d}{dx}(4x^3) = 12x^2$$

Let us find the second derivative of the function. First, we differentiate once:
$$f'(x) = e^x$$
Differentiating again gives our second derivative:
$$f''(x) = e^x$$
Since the natural exponential function $e^x$ is strictly positive for all real values of $x$, we have $f''(x) > 0$.

Therefore, the curve is concave upward on the entire domain.

A Point of Inflection is a point on a continuous curve where the direction of the curvature changes.

At this point, the graph transitions from being concave upward to concave downward, or from concave downward to concave upward. Mathematically, this requires the second derivative $f''(x)$ to equal zero (or be undefined) and change its algebraic sign.

Let us differentiate the function with respect to $x$:
$$f'(x) = 2x$$
To find stationary points, we set this equal to zero:
$$2x = 0 \implies x = 0$$
Now, let us verify using the second derivative:
$$f''(x) = 2$$
Since $f''(0) = 2 > 0$, the point represents a local minimum. Therefore, the function has a minimum at $x = 0$.

Geometrically, the equation $f'(x) = 0$ means that the slope of the tangent line to the curve at that point is zero.

This indicates that the tangent line is completely flat and horizontal (parallel to the x-axis). Such points represent peaks, valleys, or flat transition points on the graph.

Let us assume that the first part of the number is $x$. Since the sum of the two parts is $20$, the second part must be $20 - x$.

Step 1 — Write the product function:
Let us denote the product of these two parts as $P(x)$:
$$P(x) = x(20 - x) = 20x - x^2$$
Step 2 — Find the critical points:
Let us differentiate the product function with respect to $x$:
$$P'(x) = 20 - 2x$$
We set the derivative to zero to find the maximum:
$$20 - 2x = 0 \implies 2x = 20 \implies x = 10$$
Step 3 — Classify using the second derivative:
Let us calculate the second derivative:
$$P''(x) = -2$$
Since $P''(10) = -2 < 0$, the turning point at $x = 10$ is a local maximum.

Step 4 — Calculate both parts:
The first part is $x = 10$, and the second part is $20 - x = 20 - 10 = 10$.

Therefore, the two parts are 10 and 10.

Let us solve this problem systematically using the Second Derivative Test.

Step 1 — Find the first derivative:
$$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 10) = 6x^2 - 30x + 36$$
Step 2 — Solve for stationary points:
We set the derivative equal to zero:
$$6x^2 - 30x + 36 = 0$$
Divide the entire equation by $6$:
$$x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2 \quad \text{and} \quad x = 3$$
Step 3 — Find the second derivative:
$$f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30$$
Step 4 — Test our critical points:
1. At the point $x = 2$:
$$f''(2) = 12(2) - 30 = 24 - 30 = -6 < 0$$
Since the value is negative, we have a local maximum at $x = 2$. Let us calculate the corresponding maximum value:
$$f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 10 = 16 - 60 + 72 + 10 = 38$$
2. At the point $x = 3$:
$$f''(3) = 12(3) - 30 = 36 - 30 = 6 > 0$$
Since the value is positive, we have a local minimum at $x = 3$. Let us calculate the corresponding minimum value:
$$f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 10 = 54 - 135 + 108 + 10 = 37$$

Let us assume that the length of the rectangle is $x$ and its width is $y$. We are given that the area $A$ is constant:
$$A = x \cdot y \implies y = \frac{A}{x}$$
Step 1 — Set up the perimeter equation:
The perimeter of a rectangle is given by $P = 2x + 2y$. Let us substitute $y = \frac{A}{x}$ to write $P$ as a function of $x$ only:
$$P(x) = 2x + 2\left(\frac{A}{x}\right) = 2x + \frac{2A}{x}$$
Step 2 — Differentiate and locate critical points:
Let us find the first derivative of the perimeter function:
$$P'(x) = 2 - \frac{2A}{x^2}$$
We set this derivative to zero to locate the minimum value:
$$2 - \frac{2A}{x^2} = 0 \implies 2 = \frac{2A}{x^2} \implies x^2 = A \implies x = \sqrt{A}$$
Step 3 — Verify the minimum using successive differentiation:
Let us find the second derivative:
$$P''(x) = \frac{4A}{x^3}$$
Since the area $A$ and length $x$ are positive dimensions, $P''(\sqrt{A}) = \frac{4A}{A^{3/2}} > 0$. This confirms we have a local minimum.

Step 4 — Calculate the width y:
$$y = \frac{A}{x} = \frac{A}{\sqrt{A}} = \sqrt{A}$$
Since $x = y = \sqrt{A}$, the rectangle has equal sides, which proves it is a square.

Let us locate where the curvature changes by analyzing the second derivative of the function.

Step 1 — Find the first and second derivatives:
$$f'(x) = 4x^3 - 12x$$
$$f''(x) = 12x^2 - 12$$
Step 2 — Locate candidate points by setting $f''(x) = 0$:
$$12x^2 - 12 = 0 \implies 12x^2 = 12 \implies x^2 = 1 \implies x = 1 \quad \text{and} \quad x = -1$$
Step 3 — Verify that concavity changes:
Let us find the third derivative:
$$f'''(x) = 24x$$
Let us evaluate this at our candidate points:
$$f'''(1) = 24(1) = 24 \neq 0$$
$$f'''(-1) = 24(-1) = -24 \neq 0$$
Since the third derivative is non-zero, the second derivative changes signs, confirming both are valid points of inflection.

Step 4 — Find the corresponding coordinates:
1. For $x = 1$:
$$f(1) = 1^4 - 6(1)^2 + 5 = 1 - 6 + 5 = 0$$
2. For $x = -1$:
$$f(-1) = (-1)^4 - 6(-1)^2 + 5 = 1 - 6 + 5 = 0$$

Therefore, the points of inflection are $(1, 0)$ and $(-1, 0)$.

Let us locate the absolute maximum and minimum values by evaluating our function at its critical points and endpoints.

Step 1 — Find the critical points:
$$f'(x) = 3x^2 - 12x + 9$$
We set this derivative to zero:
$$3x^2 - 12x + 9 = 0 \implies x^2 - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0 \implies x = 1 \quad \text{and} \quad x = 3$$
Both points lie within our closed interval $[0, 4]$.

Step 2 — Evaluate the function at all critical points and boundaries:
1. At the boundary $x = 0$:
$$f(0) = 0^3 - 6(0)^2 + 9(0) + 2 = 2$$
2. At the critical point $x = 1$:
$$f(1) = 1^3 - 6(1)^2 + 9(1) + 2 = 1 - 6 + 9 + 2 = 6$$
3. At the critical point $x = 3$:
$$f(3) = 3^3 - 6(3)^2 + 9(3) + 2 = 27 - 54 + 27 + 2 = 2$$
4. At the boundary $x = 4$:
$$f(4) = 4^3 - 6(4)^2 + 9(4) + 2 = 64 - 96 + 36 + 2 = 6$$

Step 3 — Compare the values:
The largest value is $6$ (occurring at $x = 1$ and $x = 4$) and the smallest value is $2$ (occurring at $x = 0$ and $x = 3$).

Therefore, the absolute maximum value is $6$ and the absolute minimum value is $2$.

Let us calculate the second derivative of our function to determine its concavity.

Step 1 — Differentiate successively:
$$f'(x) = 6x^2 - 24x + 18$$
$$f''(x) = 12x - 24$$
Step 2 — Find where the concavity is zero:
$$12x - 24 = 0 \implies 12x = 24 \implies x = 2$$
Step 3 — Test the intervals:
This divides the real line into two intervals: $(-\infty, 2)$ and $(2, \infty)$.

1. For the interval $(-\infty, 2)$, let us choose a test point $x = 0$:
$$f''(0) = 12(0) - 24 = -24 < 0$$
Since the second derivative is negative, the curve is concave downward on $(-\infty, 2)$.

2. For the interval $(2, \infty)$, let us choose a test point $x = 3$:
$$f''(3) = 12(3) - 24 = 36 - 24 = 12 > 0$$
Since the second derivative is positive, the curve is concave upward on $(2, \infty)$.

Let us assume that the first positive number is $x$. Since their sum is $16$, the second number must be $16 - x$.

Step 1 — Set up the sum of squares function:
Let us denote the sum of their squares as $S(x)$:
$$S(x) = x^2 + (16 - x)^2$$
Let us expand this equation:
$$S(x) = x^2 + 256 - 32x + x^2 = 2x^2 - 32x + 256$$
Step 2 — Find critical points:
Let us differentiate with respect to $x$:
$$S'(x) = 4x - 32$$
We set this derivative equal to zero to find the minimum:
$$4x - 32 = 0 \implies 4x = 32 \implies x = 8$$
Step 3 — Classify using the second derivative test:
$$S''(x) = 4$$
Since the second derivative is a positive constant $4 > 0$, the turning point at $x = 8$ represents a local minimum.

Step 4 — Calculate both values:
The first number is $x = 8$, and the second number is $16 - x = 16 - 8 = 8$.

Therefore, the two positive numbers are 8 and 8.

Let us assume the length of the rectangle is $x$ and its width is $y$. The perimeter of the rectangle is given as $20\text{ cm}$:
$$2(x + y) = 20 \implies x + y = 10 \implies y = 10 - x$$
Step 1 — Set up the area function:
Let us express the area $A$ in terms of $x$:
$$A(x) = x \cdot y = x(10 - x) = 10x - x^2$$
Step 2 — Differentiate and locate critical points:
Let us differentiate the area function:
$$A'(x) = 10 - 2x$$
We set the derivative to zero:
$$10 - 2x = 0 \implies 2x = 10 \implies x = 5\text{ cm}$$
Step 3 — Verify using the second derivative:
$$A''(x) = -2$$
Since $A''(5) = -2 < 0$, the turning point at $x = 5$ represents a local maximum.

Step 4 — Calculate the width:
$$y = 10 - x = 10 - 5 = 5\text{ cm}$$

Therefore, the dimensions of the rectangle for maximum area are $5\text{ cm}$ by $5\text{ cm}$.

Let us find the first derivative of the given function to identify critical points:
$$f'(x) = 3x^2 - 6x + 3$$
We set the derivative equal to zero:
$$3x^2 - 6x + 3 = 0 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1$$
This gives us a single critical point at $x = 1$. Let us apply the Second Derivative Test:
$$f''(x) = 6x - 6$$
Substituting $x = 1$ gives:
$$f''(1) = 6(1) - 6 = 0$$
Since the second derivative is zero, the Second Derivative Test is inconclusive. Let us use the First Derivative Test and examine the sign of $f'(x)$ immediately around $x = 1$:

1. For $x < 1$ (e.g., $x = 0$):
$$f'(0) = 3(0)^2 - 6(0) + 3 = 3 > 0$$
2. For $x > 1$ (e.g., $x = 2$):
$$f'(2) = 3(2)^2 - 6(2) + 3 = 12 - 12 + 3 = 3 > 0$$

Notice that the sign of $f'(x)$ is positive on both sides of $x = 1$. Since the slope does not change its sign, the point $x = 1$ is a point of inflection rather than a maximum or minimum. Therefore, the function has neither a maximum nor a minimum.

Let us consider a car driving along a straight road. We can represent the position of the car at any time $t$ as a function $s(t)$.

If we look at a change in time $\Delta t$, the average velocity over that interval is:
$$\text{Average Velocity} = \frac{\Delta s}{\Delta t} = \frac{s(t + \Delta t) - s(t)}{\Delta t}$$
Now, if we shrink the time interval $\Delta t$ down to zero, we find the velocity at one exact instant:
$$\text{Instantaneous Velocity } v(t) = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt} = s'(t)$$
Therefore, physically, the first derivative of position with respect to time represents the instantaneous velocity of the car. If we differentiate a second time, the rate of change of velocity gives the instantaneous acceleration:
$$a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} = s''(t)$$