Understanding Maximum, Minimum Values, slope and stationary points

“Maximum” and “minimum” are terms used to describe the highest and lowest possible values of something.

Maximum refers to the greatest or highest amount, number, or level that is possible. For example, the maximum speed of a car is the fastest it can go.
Minimum refers to the smallest or lowest amount, number, or level that is possible. For instance, the minimum temperature in winter is the coldest it gets.

Some real-life examples of maximum and minimum:

Temperature: The highest recorded temperature on Earth was 56.7°C (134°F) in Death Valley, California, while the lowest was −89.2°C (−128.6°F) in Antarctica.
Speed Limits: The maximum speed limit on highways varies by country—some German autobahns have no limit, while others cap at around 120–130 km/h (75–80 mph). Residential areas often have a minimum speed limit to keep traffic moving safely.
Human Age: The oldest verified person lived to 122 years (Jeanne Calment of France), while newborn babies start at zero years!
Height: The tallest mountain is Mount Everest (8,848m / 29,029 ft), while the lowest point on land is the Dead Sea shore (~430m / ~1,412 ft below sea level).

Maximum and minimum are crucial in economics, shaping policies, market behavior, and financial decisions. Here are some key applications:

Price Ceilings & Floors: Governments sometimes set a maximum price (price ceiling) to make essentials like housing or medicine affordable. Conversely, a minimum price (price floor) ensures fair wages—like minimum wage laws.
Supply & Demand: Businesses analyze the maximum and minimum demand for goods to optimize production. A company might produce minimum quantities of a product when demand is low and maximize production when demand spikes.
Profit & Loss: Companies aim to maximize profit by increasing revenue and reducing costs. On the flip side, they try to minimize losses during economic downturns.
Interest Rates: Central banks adjust interest rates to maximize economic growth or minimize inflation. Lower rates encourage spending, while higher rates control excessive borrowing.
Resource Allocation: Governments and companies aim to minimize waste and maximize efficiency in distributing resources, whether it’s food, energy, or labor.

The slope of a function describes how steep it is and how it changes as you move along the graph. In mathematical terms, it’s the rate at which the function’s value increases or decreases with respect to changes in the input (usually (x)). So, it simply the derivative of the function at x. In other words, the derivative ( f'(x) ) of a function ( f(x) ) gives the slope at any given (x).

A positive slope means the function is increasing (going uphill).
A negative slope means the function is decreasing (going downhill).
A zero slope means the function is flat at that point (no change in ( y )).

Example:

For the linear function ( y = 2x + 3 ), the slope is 2, meaning for every unit increase in (x), (y) increases by 2.

In summary, the derivative of a function at a certain point gives you the slope of the tangent line to the function’s graph at that specific point. In essence, it tells you the instantaneous rate of change of the function.

Here’s how to find it:

Start with a function: Let’s say you have a function $f(x)$ .
Compute the derivative: Find the derivative of the function, denoted as $f'(x)$ or $\frac{dy}{dx}$ . This new function, $f'(x)$ , is a formula that gives the slope of $f(x)$ at any value of $x$ .
Evaluate the derivative at a specific point: To find the slope at a particular point, say $x = a$ , you substitute $a$ into the derivative: $m = f'(a)$ .

Here are a few examples:

Example 1: A Simple Quadratic Function

Let the function be $f(x) = x^2$ .

Compute the derivative:
Using the power rule (which states that if $f(x) = x^n$ , then $f'(x) = nx^{n-1}$ ):
$f'(x) = 2x^{2-1} = 2x$
Interpretation:
The derivative $f'(x) = 2x$ tells us the slope of the tangent line to the graph of $f(x) = x^2$ at any point $x$ .
Find the slope at a specific point, e.g., $x = 3$ :
Substitute $x = 3$ into the derivative $f'(x)$ :
Slope $m = f'(3) = 2(3) = 6$ So, the slope of the tangent line to the curve $y = x^2$ at the point where $x = 3$ (which is the point $(3, 9)$ ) is 6. Find the slope at another point, e.g., $x = -1$ :
Slope $m = f'(-1) = 2(-1) = -2$
At $x = -1$ (the point $(-1, 1)$ ), the slope of the tangent line is -2.

Example 2: A Cubic Function

Let the function be $g(x) = x^3 - 4x + 2$ .

Compute the derivative:
We use the power rule for each term and the sum/difference rule (the derivative of a sum/difference is the sum/difference of the derivatives). The derivative of a constant (like 2) is 0.
$g'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(4x) + \frac{d}{dx}(2)$
$g'(x) = 3x^{3-1} - 4x^{1-1} + 0$
$g'(x) = 3x^2 - 4x^0$
$g'(x) = 3x^2 - 4$ (since $x^0 = 1$ )
Interpretation:
The derivative $g'(x) = 3x^2 - 4$ gives the slope of the tangent line to $g(x) = x^3 - 4x + 2$ at any point $x$ .
Find the slope at a specific point, e.g., $x = 2$ :
Substitute $x = 2$ into $g'(x)$ :
Slope $m = g'(2) = 3(2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8$ So, the slope of the tangent line to $y = x^3 - 4x + 2$ at $x = 2$ (the point $(2, 2^3 - 4(2) + 2) = (2, 8 - 8 + 2) = (2, 2)$ ) is 8.

A stationary point of a differentiable function of a single variable is a point on the graph of the function where the derivative is zero. In simpler terms, it’s a point where the slope of the tangent line to the curve is horizontal (flat).

Mathematically, for a function $f(x)$ , a point $x_0$ is a stationary point if:
$f'(x_0) = 0$

Why are stationary points important?
Stationary points are crucial in calculus and function analysis because they often correspond to interesting features of the graph, such as:

Local Maximum: The highest point in its immediate neighborhood (like the peak of a hill). The function changes from increasing to decreasing at this point.
Local Minimum: The lowest point in its immediate neighborhood (like the bottom of a valley). The function changes from decreasing to increasing at this point.
Horizontal Point of Inflection (or Saddle Point for 2D graphs): A point where the function changes its concavity (from curving upwards to curving downwards, or vice versa), and the tangent is horizontal. The function does not change its general direction of increasing or decreasing, or it briefly flattens out before continuing in the same direction.

How to find stationary points:

Differentiate the function: Find the first derivative of the function, $f'(x)$ .
Set the derivative to zero: Solve the equation $f'(x) = 0$ for $x$ . The solutions for $x$ are the x-coordinates of the stationary points.
Find the y-coordinates (optional but good for plotting): Substitute these x-values back into the original function $f(x)$ to find the corresponding y-coordinates.

How to classify stationary points:

Once you’ve found a stationary point, you can classify it using one of these tests:

First Derivative Test:
- Examine the sign of the derivative on either side of the stationary point .
  - If $f'(x)$ changes from positive (+) to negative (-), then $x_0$ is a local maximum.
  - If $f'(x)$ changes from negative (-) to positive (+), then $x_0$ is a local minimum.
  - If $f'(x)$ does not change sign (e.g., positive to positive, or negative to negative), then $x_0$ is a horizontal point of inflection.
Second Derivative Test:
- Calculate the second derivative of the function, $f''(x)$ .
- Evaluate at the stationary point .
  - If $f''(x_0) < 0$ (negative), then $x_0$ is a local maximum.
  - If $f''(x_0) > 0$ (positive), then $x_0$ is a local minimum.
  - If $f''(x_0) = 0$ , the test is inconclusive. You must use the first derivative test or examine higher-order derivatives. This case might indicate a point of inflection.

Example:

Let’s find and classify the stationary points of the function $f(x) = x^3 - 3x + 2$ .

Differentiate the function:
$f'(x) = 3x^2 - 3$
Set the derivative to zero:
$3x^2 - 3 = 0$
$3x^2 = 3$
$x^2 = 1$
$x = 1$ or $x = -1$
So, the x-coordinates of the stationary points are $1$ and $-1$ .
Classify using the Second Derivative Test:
First, find the second derivative:
$f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x$ Now, evaluate $f''(x)$ at $x = 1$ :
$f''(1) = 6(1) = 6$
Since $f''(1) > 0$ , the point at $x = 1$ is a local minimum.
The y-coordinate is $f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$ . So, $(1, 0)$ is a local minimum. Evaluate $f''(x)$ at $x = -1$ :
$f''(-1) = 6(-1) = -6$
Since $f''(-1) < 0$ , the point at $x = -1$ is a local maximum.
The y-coordinate is $f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$ . So, $(-1, 4)$ is a local maximum.

In summary, stationary points are fundamental for understanding the shape and behavior of functions. They are the “flat spots” on a curve where many important local changes occur.

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