Unveiling Local Extrema: A Comprehensive Guide

Oct 28, 2025 by Admin 47 views

Hey guys! Today, we're diving deep into the fascinating world of calculus to find the local extrema of some cool functions. Local extrema, you ask? Think of them as the peaks and valleys of a function's graph – the points where the function reaches a maximum or minimum value in a specific region. We'll be using derivatives, critical points, and a little bit of intuition to hunt down these special locations. Buckle up, because we're about to embark on an exciting mathematical adventure!

(1) Exploring the Peaks and Valleys of f(x) = sin³(x) + cos³(x)

Alright, let's start with our first function: f(x) = sin³(x) + cos³(x), where x ranges from 0 to 2π. Our goal here is to find the local extrema. To do this, we need to find where the function's derivative is either equal to zero or undefined. These spots are called critical points, and they're prime locations for local maxima or minima. So, first things first, let's find the derivative, f'(x).

To find f'(x), we'll use the chain rule. The derivative of sin³(x) is 3sin²(x)cos(x), and the derivative of cos³(x) is -3cos²(x)sin(x). Putting it all together, we have f'(x) = 3sin²(x)cos(x) - 3cos²(x)sin(x). Now, let's simplify this mess. We can factor out 3sin(x)cos(x), giving us f'(x) = 3sin(x)cos(x)[sin(x) - cos(x)].

Next, we need to find the critical points. This means setting f'(x) = 0 and solving for x. So, we have 3sin(x)cos(x)[sin(x) - cos(x)] = 0. This equation is satisfied when either sin(x) = 0, cos(x) = 0, or sin(x) = cos(x).

Let's break these down:

sin(x) = 0: This occurs when x = 0, π, and 2π.
cos(x) = 0: This occurs when x = π/2 and 3π/2.
sin(x) = cos(x): This occurs when x = π/4 and 5π/4 (within the interval [0, 2π]).

So, our critical points are x = 0, π/4, π/2, 5π/4, 3π/2, and 2π. Now, we need to determine whether these points are local maxima, local minima, or neither. We can use the first derivative test for this. We'll check the sign of f'(x) in the intervals between our critical points. This will tell us whether the function is increasing or decreasing.

Let's analyze the intervals:

(0, π/4): Choose x = π/8. f'(π/8) > 0, so the function is increasing.
(π/4, π/2): Choose x = π/3. f'(π/3) < 0, so the function is decreasing.
(π/2, 5π/4): Choose x = π. f'(π) > 0, so the function is increasing.
(5π/4, 3π/2): Choose x = 7π/4. f'(7π/4) < 0, so the function is decreasing.
(3π/2, 2π): Choose x = 11π/4. f'(11π/4) > 0, so the function is increasing.

From our analysis, we can conclude that we have:

Local maximum at x = π/4 and x = 2π.
Local minimum at x = π/2 and x = 5π/4.

Remember to plug these x values back into the original function f(x) to find the actual y values of the local extrema. The point (0, 1) and (2π, 1) are local maximum, the point (π/4, 0.354) is a local maximum, the point (π/2, 0) is a local minimum, and the point (5π/4, -0.354) is a local minimum, and the point (3π/2, 0) is a local minimum.

(2) Navigating the Absolute Value: f(x) = |x(x² - 4)|

Alright, let's shift gears and tackle f(x) = |x(x² - 4)|. The absolute value signs add a little twist to our journey. Remember, the absolute value of a number is its distance from zero, so it always results in a non-negative value. Because of this, our function could have sharp turns where the derivative isn't defined. We need to be especially mindful of where the expression inside the absolute value changes sign.

First, let's figure out where the expression inside the absolute value, x(x² - 4), equals zero. This is where x = 0, x = 2, and x = -2. These points are crucial because they're where the function might have non-differentiable points. We can rewrite the function as follows:

f(x) = |x(x² - 4)| = |x(x - 2)(x + 2)|

Now, let's consider the intervals determined by these critical points:

For x < -2, x(x² - 4) is negative, so f(x) = -x(x² - 4) = -x³ + 4x.
For -2 < x < 0, x(x² - 4) is positive, so f(x) = x(x² - 4) = x³ - 4x.
For 0 < x < 2, x(x² - 4) is negative, so f(x) = -x(x² - 4) = -x³ + 4x.
For x > 2, x(x² - 4) is positive, so f(x) = x(x² - 4) = x³ - 4x.

Now, let's find the derivative for each of these cases:

For x < -2: f'(x) = -3x² + 4.
For -2 < x < 0: f'(x) = 3x² - 4.
For 0 < x < 2: f'(x) = -3x² + 4.
For x > 2: f'(x) = 3x² - 4.

Let's find the critical points for the derivative being zero. Setting the derivates above equal to zero, we get the values of x equal to +2/sqrt(3) or -2/sqrt(3).

For x < -2: f'(x) = 0 when x = -2/sqrt(3) but this is not in the range.
For -2 < x < 0: f'(x) = 0 when x = -2/sqrt(3) and this is in the range.
For 0 < x < 2: f'(x) = 0 when x = 2/sqrt(3) and this is in the range.
For x > 2: f'(x) = 0 when x = 2/sqrt(3) but this is not in the range.

Now we check for local extrema by checking the sign of the derivate in each region:

For x < -2, we have f'(-3) = -23. The function is decreasing.
For -2 < x < -2/sqrt(3), we have f'(-1.5) = 2.75. The function is increasing.
For -2/sqrt(3) < x < 0, we have f'(-1) = -1. The function is decreasing.
For 0 < x < 2/sqrt(3), we have f'(1) = 1. The function is increasing.
For 2/sqrt(3) < x < 2, we have f'(1.5) = -2.75. The function is decreasing.
For x > 2, we have f'(3) = 23. The function is increasing.

So the local extrema is when x = -2/sqrt(3), x = 0, and x = 2/sqrt(3). Remember to plug these values back into the original function to find the corresponding y values. The local minimum is (0, 0), the local maximum is (-2/sqrt(3), 5.196), and the local maximum is (2/sqrt(3), 5.196).

(3) Unraveling f(x) = x(x² + 1) / (x⁴ - x² + 1)

Let's move on to f(x) = x(x² + 1) / (x⁴ - x² + 1). This function involves a rational expression, which means we need to be extra cautious about the denominator. Remember, division by zero is a big no-no, so we'll need to make sure the denominator never equals zero. In this case, x⁴ - x² + 1 is never equal to zero. Let's find the derivative, f'(x), and then find critical points.

To find the derivative of this function, we'll use the quotient rule. The quotient rule states that if we have a function f(x) = u(x) / v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x)) / v(x)². Let's apply this to our function.

Let u(x) = x(x² + 1) = x³ + x, so u'(x) = 3x² + 1.
Let v(x) = x⁴ - x² + 1, so v'(x) = 4x³ - 2x.

Now, plug these into the quotient rule formula:

f'(x) = [(3x² + 1)(x⁴ - x² + 1) - (x³ + x)(4x³ - 2x)] / (x⁴ - x² + 1)²

Let's simplify that numerator. Expanding the terms, we get:

f'(x) = [3x⁶ - 3x⁴ + 3x² + x⁴ - x² + 1 - (4x⁶ - 2x⁴ + 4x² - 2x²)] / (x⁴ - x² + 1)²

f'(x) = [3x⁶ - 2x⁴ + 2x² + 1 - 4x⁶ + 2x⁴ - 2x²] / (x⁴ - x² + 1)²

f'(x) = [-x⁶ + 1] / (x⁴ - x² + 1)²

Now, to find the critical points, we set f'(x) = 0. This means the numerator must equal zero: -x⁶ + 1 = 0.

Solving for x, we get x⁶ = 1, which has solutions x = 1 and x = -1. These are our critical points.

Let's analyze the intervals determined by the critical points:

For x < -1, let's test x = -2. f'(-2) = -63/225. The function is decreasing.
For -1 < x < 1, let's test x = 0. f'(0) = 1. The function is increasing.
For x > 1, let's test x = 2. f'(2) = -63/225. The function is decreasing.

Therefore we have a local minimum at x = -1 and a local maximum at x = 1. Again, remember to plug these x-values back into the original function to find the corresponding y-values. The local minimum is (-1, -0.667), and the local maximum is (1, 0.667).

(4) Investigating g(t) = |t² - 4t + 1| on [0, 5]

Alright, let's wrap things up with g(t) = |t² - 4t + 1|, restricted to the interval [0, 5]. We're back to dealing with an absolute value function, so remember to pay attention to where the expression inside the absolute value changes sign. The absolute value function can introduce potential sharp turns in our graph, where the derivative might be undefined.

First, let's find the values of t where the expression inside the absolute value, t² - 4t + 1, equals zero. We can use the quadratic formula to solve for t:

t = (-b ± √(b² - 4ac)) / 2a

In our case, a = 1, b = -4, and c = 1. Plugging these values into the formula, we get:

t = (4 ± √((-4)² - 4 * 1 * 1)) / (2 * 1) t = (4 ± √(16 - 4)) / 2 t = (4 ± √12) / 2 t = (4 ± 2√3) / 2 t = 2 ± √3

So, our critical points are t = 2 + √3 ≈ 3.732 and t = 2 - √3 ≈ 0.268. Both of these values fall within our interval [0, 5]. Remember to include the end points of the interval as well, which are t = 0 and t = 5.

Now, let's analyze the intervals determined by these critical points and endpoints:

For 0 ≤ t < 2 - √3, t² - 4t + 1 is positive, so g(t) = t² - 4t + 1.
For 2 - √3 < t < 2 + √3, t² - 4t + 1 is negative, so g(t) = -(t² - 4t + 1) = -t² + 4t - 1.
For 2 + √3 < t ≤ 5, t² - 4t + 1 is positive, so g(t) = t² - 4t + 1.

Now let us find the derivate. For the interval 0 ≤ t < 2 - √3, g'(t) = 2t - 4. Setting it to 0, we get t = 2 which is not in the range.

For the interval 2 - √3 < t < 2 + √3, g'(t) = -2t + 4. Setting it to 0, we get t = 2 which is in the range.

For the interval 2 + √3 < t ≤ 5, g'(t) = 2t - 4. Setting it to 0, we get t = 2 which is not in the range.

Let's analyze the intervals determined by these critical points and endpoints:

For 0 ≤ t < 0.268, g'(t) = 2t - 4. g'(0) = -4, which means it is decreasing.
For 0.268 < t < 2, g'(t) = -2t + 4. g'(1) = 2, which means it is increasing.
For 2 < t < 3.732, g'(t) = -2t + 4. g'(3) = -2, which means it is decreasing.
For 3.732 < t ≤ 5, g'(t) = 2t - 4. g'(4) = 4, which means it is increasing.

So, we have critical points at t = 0, 2 - √3, 2, 2 + √3, and 5.

Local minimum at t = 0 and t = 2 + √3. The points are (0, 1) and (2 + √3, 0).
Local maximum at t = 2 - √3, and t = 5. The points are (2 - √3, 0) and (5, 6).
Local minimum at t = 2. The point is (2, 3).

And there you have it, guys! We've successfully navigated through all the function, finding their local extrema. Remember, practice makes perfect, so keep working through problems, and you'll become a local extrema pro in no time! Keep exploring and having fun with math! Bye!