Finding Critical Numbers For F(x) = X^(4/5)(x-3)^2

Hey guys! Today, we're diving into a fun math problem: finding the critical numbers of the function $F(x) = x^{4/5}(x-3)^2$. This might sound intimidating, but don't worry, we'll break it down step by step. Critical numbers are super important in calculus because they help us understand where a function might have local maxima, local minima, or saddle points. So, let's get started!

Understanding Critical Numbers

Before we jump into the calculations, let's quickly recap what critical numbers actually are. A critical number of a function $F(x)$ is a value $c$ in the domain of $F$ where either $F'(c) = 0$ or $F'(c)$ does not exist. In simpler terms, these are the points where the function's slope is either zero (meaning a horizontal tangent line) or undefined (meaning a vertical tangent line or a sharp turn). Finding these points is crucial for analyzing the behavior of a function, such as identifying its increasing and decreasing intervals, and locating its local extrema.

To really nail this down, think about what the derivative, $F'(x)$, represents. It's the slope of the tangent line to the function at any given point. When $F'(x) = 0$, the tangent line is horizontal, indicating a potential peak or valley in the function's graph. When $F'(x)$ is undefined, it suggests the function has a point where it's not smooth, like a cusp or a vertical tangent. These are all key spots to investigate when sketching the graph or understanding the function's behavior.

Now, why are critical numbers so important? Well, they act as the signposts for the function's journey. Imagine you're on a roller coaster; the critical points are like the tops of the hills and the bottoms of the valleys. These are the points where the roller coaster might change direction, and similarly, these are the points where the function might switch from increasing to decreasing, or vice versa. By finding these critical numbers, we can create a sort of roadmap for the function's behavior, allowing us to sketch its graph more accurately and predict its maximum and minimum values.

In the context of real-world applications, critical numbers pop up everywhere. For instance, in optimization problems, we often need to find the maximum or minimum value of a function. This could be anything from maximizing profit to minimizing cost. Critical numbers are our go-to tools for solving these problems. Similarly, in physics, they can help us find the points where velocity is zero (like when a ball reaches the peak of its trajectory) or where acceleration changes direction. So, understanding how to find and interpret critical numbers is not just a theoretical exercise; it's a practical skill that has wide-ranging applications.

Step 1: Finding the Derivative F'(x)

The first step in finding the critical numbers is to find the derivative of the function. For $F(x) = x^{4/5}(x-3)^2$, we'll need to use the product rule and the chain rule. Remember, the product rule states that if we have a function $F(x) = u(x)v(x)$, then $F'(x) = u'(x)v(x) + u(x)v'(x)$. The chain rule helps us differentiate composite functions, telling us that if $F(x) = g(h(x))$, then $F'(x) = g'(h(x))h'(x)$.

Let's identify our $u(x)$ and $v(x)$ in our function. We can set $u(x) = x^{4/5}$ and $v(x) = (x-3)^2$. Now, we need to find the derivatives of $u(x)$ and $v(x)$.

For $u(x) = x^{4/5}$, we use the power rule, which says that if $u(x) = x^n$, then $u'(x) = nx^{n-1}$. Applying this, we get:

u'(x) = rac{4}{5}x^{(4/5) - 1} = rac{4}{5}x^{-1/5}

Next, we find the derivative of $v(x) = (x-3)^2$. Here, we need the chain rule. We can think of $v(x)$ as a composite function where the outer function is squaring and the inner function is $(x-3)$. The derivative of the outer function (squaring) is $2$ times the inside, and the derivative of the inner function $(x-3)$ is just $1$. So, applying the chain rule:

$v'(x) = 2(x-3)(1) = 2(x-3)$

Now that we have $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$, we can plug them into the product rule formula:

$F'(x) = u'(x)v(x) + u(x)v'(x)$

F'(x) = rac{4}{5}x^{-1/5}(x-3)^2 + x^{4/5}(2(x-3))

This is our derivative, but it looks a bit messy. To make it easier to work with, we'll simplify it in the next step.

Step 2: Simplifying the Derivative

Okay, guys, let's take that derivative we found and clean it up a bit. Simplifying the derivative $F'(x)$ will make it much easier to find the critical numbers. We have:

F'(x) = rac{4}{5}x^{-1/5}(x-3)^2 + x^{4/5}(2(x-3))

The goal here is to combine these two terms into a single fraction. To do that, we'll need a common factor. Notice that both terms have a factor of $(x-3)$ and a power of $x$. Let's factor out the lowest power of $x$, which is $x^{-1/5}$, and the lowest power of $(x-3)$, which is $(x-3)$. Factoring these out, we get:

F'(x) = x^{-1/5}(x-3) igg[rac{4}{5}(x-3) + 2xigg]

Now, let's focus on simplifying the expression inside the brackets:

rac{4}{5}(x-3) + 2x = rac{4}{5}x - rac{12}{5} + 2x

To combine these terms, we need a common denominator. Let's rewrite $2x$ as rac{10x}{5}:

rac{4}{5}x - rac{12}{5} + rac{10x}{5} = rac{14x}{5} - rac{12}{5}

Now we can write this as a single fraction:

rac{14x - 12}{5}

Putting it all back together, our simplified derivative is:

F'(x) = x^{-1/5}(x-3)rac{14x - 12}{5}

We can also rewrite $x^{-1/5}$ as rac{1}{x^{1/5}}, so:

F'(x) = rac{(x-3)(14x - 12)}{5x^{1/5}}

This simplified form is much easier to work with. Now we can move on to finding the critical numbers by setting the derivative equal to zero and looking for points where it's undefined.

Step 3: Finding Where F'(x) = 0

Alright, let's find the critical numbers where the derivative is equal to zero. This means we need to solve the equation $F'(x) = 0$. Remember our simplified derivative:

F'(x) = rac{(x-3)(14x - 12)}{5x^{1/5}}

A fraction is equal to zero when its numerator is equal to zero. So, we need to solve:

$(x-3)(14x - 12) = 0$

This gives us two factors to consider:

1. $x - 3 = 0$
2. $14x - 12 = 0$

Solving the first equation, we get:

$x = 3$

Solving the second equation:

$14x = 12$

x = rac{12}{14} = rac{6}{7}

So, we have two potential critical numbers where the derivative is zero: $x = 3$ and x = rac{6}{7}. These are the points where the function has a horizontal tangent line, which could indicate a local maximum or minimum.

Step 4: Finding Where F'(x) is Undefined

Now, let's find the critical numbers where the derivative is undefined. Looking back at our simplified derivative:

F'(x) = rac{(x-3)(14x - 12)}{5x^{1/5}}

A fraction is undefined when its denominator is equal to zero. So, we need to find where:

$5x^{1/5} = 0$

Dividing both sides by 5, we get:

$x^{1/5} = 0$

Raising both sides to the power of 5, we find:

$x = 0$

So, $x = 0$ is another potential critical number. At this point, the derivative is undefined, which could indicate a vertical tangent or a sharp corner in the graph of the function.

Step 5: Check if Critical Points are in the Domain of F(x)

Before we finalize our list of critical numbers, it's essential to make sure they're actually in the domain of the original function, $F(x) = x^{4/5}(x-3)^2$. The domain is the set of all possible $x$-values that we can plug into the function without causing any issues (like dividing by zero or taking the square root of a negative number).

Our function has two parts: $x^{4/5}$ and $(x-3)^2$. The $(x-3)^2$ part is a polynomial, and polynomials are defined for all real numbers. So, no issues there.

The $x^{4/5}$ part can be rewritten as $( ext{5th root of }x)^4$. Taking the fifth root of a number is perfectly fine for any real number, whether it's positive, negative, or zero. And raising that result to the fourth power also doesn't cause any problems. Therefore, $x^{4/5}$ is also defined for all real numbers.

Since both parts of the function are defined for all real numbers, the domain of $F(x)$ is all real numbers. This means that all the critical numbers we found—$x = 0$, x = rac{6}{7}, and $x = 3$—are indeed in the domain of $F(x)$.

Why is this step so important? Imagine we had a critical number that wasn't in the domain. That would mean the function isn't even defined at that point, so it couldn't possibly be a local max, local min, or anything interesting. Checking the domain ensures that our critical numbers are legitimate candidates for these special points.

Final Answer

Alright, guys, we've done it! We've found all the critical numbers for the function $F(x) = x^{4/5}(x-3)^2$. Let's recap what we did:

1. Found the derivative $F'(x)$ using the product rule and chain rule.
2. Simplified the derivative to make it easier to work with.
3. Found where $F'(x) = 0$ by setting the numerator of the derivative equal to zero.
4. Found where $F'(x)$ is undefined by setting the denominator of the derivative equal to zero.
5. Checked that all critical numbers are in the domain of the original function.

Our critical numbers are:

x = 0, rac{6}{7}, 3

These are the points where the function might have local maxima, local minima, or points of inflection. To determine exactly what's happening at these points, we could use the first derivative test or the second derivative test, but that's a topic for another day. For now, give yourselves a pat on the back for successfully navigating this problem! You've tackled a challenging calculus question, and that's something to be proud of. Keep practicing, and you'll become even more confident in your calculus skills.