Find The Absolute Max Of F(x)=x^4+8x^3+18x^2+3 On [-4,1]

Hey math whizzes! Today, we're diving into a classic calculus problem: finding the absolute maximum value of a function on a closed interval. It's like a treasure hunt, but instead of gold, we're looking for the highest point on the function's graph within a specific range. Our mission, should we choose to accept it, is to determine the absolute maximum value of the function $f(x)=x^4+8 x^3+18 x^2+3$ on the interval $[-4,1]$. This might seem a bit daunting with all those $x$ terms, but trust me, with a systematic approach, we'll conquer it. We'll be using the power of derivatives to find critical points, and then comparing function values at these points and the interval's endpoints. So, grab your thinking caps, and let's get started on this mathematical adventure! We'll break down each step to make sure everyone's on board, whether you're a calculus newbie or a seasoned pro. Remember, practice makes perfect, and understanding these fundamental concepts will set you up for success in all sorts of calculus challenges.

Understanding the Absolute Maximum Value Concept

Alright guys, let's get our heads around what we're actually trying to find. The absolute maximum value of a function on a closed interval is simply the largest output value the function produces for any input within that specific interval. Think of it like this: imagine you're hiking on a mountain range, and you're restricted to a particular path (our closed interval). The absolute maximum value is the highest altitude you reach during that hike. It's not just any peak; it's the highest point on your entire journey. For a continuous function on a closed interval, the Extreme Value Theorem guarantees that such an absolute maximum (and an absolute minimum) must exist. This is super important because it tells us we don't have to worry about the function doing something wild and avoiding its highest value. The candidates for this absolute maximum will always be found at either the critical points of the function within the interval or at the endpoints of the interval. Critical points are those special places where the function's derivative is either zero or undefined. These are often where the function changes direction, potentially reaching a peak or a valley. The endpoints, on the other hand, are the literal boundaries of our interval. By evaluating the function at all these potential locations, we can confidently identify the single highest value. It's like checking all the highest spots on your map to see which one is truly the summit. So, our strategy will be to find these critical points, check the function's value at them if they're inside our interval, and then compare those values with the function's values at the interval's start and end. The biggest number we get from all these calculations will be our prize – the absolute maximum value!

Step 1: Finding the Critical Points

Now, for the nitty-gritty part: finding those all-important critical points. These are the points where the function's behavior might change, giving us potential peaks for our absolute maximum. To find them, we need to get our hands dirty with the derivative of $f(x)$. Our function is $f(x)=x^4+8 x^3+18 x^2+3$. Let's find its derivative, $f'(x)$, using the power rule, which states that the derivative of $x^n$ is $nx^{n-1}$.

So, the derivative of $x^4$ is $4x^3$. The derivative of $8x^3$ is $8 imes 3x^{3-1} = 24x^2$. The derivative of $18x^2$ is $18 imes 2x^{2-1} = 36x$. And the derivative of a constant, 3, is just 0.

Putting it all together, our derivative is $f'(x) = 4x^3 + 24x^2 + 36x$.

Next, we need to find where this derivative is either equal to zero or undefined. Since $f'(x)$ is a polynomial, it's defined for all real numbers, so we only need to worry about where $f'(x) = 0$.

Let's set the derivative to zero: $4x^3 + 24x^2 + 36x = 0$.

To solve this, we can factor out the greatest common factor, which is $4x$: $4x(x^2 + 6x + 9) = 0$.

Now, we need to factor the quadratic expression inside the parentheses: $x^2 + 6x + 9$. This is a perfect square trinomial, which factors into $(x+3)^2$.

So, our equation becomes: $4x(x+3)^2 = 0$.

For this product to be zero, at least one of the factors must be zero. This gives us two possibilities:

1. $4x = 0

ightarrow x = 0$ 2. $(x+3)^2 = 0

ightarrow x+3 = 0

ightarrow x = -3$

These are our critical points: $x = 0$ and $x = -3$. These are the locations where the function might have a local maximum or minimum. Remember, these are potential candidates for our absolute maximum, but we also need to consider the interval's endpoints.

Step 2: Evaluate the Function at Critical Points and Endpoints

Alright, we've found our critical points ($x=0$ and $x=-3$). Now, we need to check if these points lie within our given interval, which is $[-4, 1]$. Luckily, both $0$ and $-3$ are indeed within this interval! So, they are valid candidates for our absolute maximum.

Our next crucial step is to evaluate the original function, $f(x)=x^4+8 x^3+18 x^2+3$, at these critical points and at the endpoints of our interval, which are $x = -4$ and $x = 1$. This is where we'll get the actual function values (the y-values) that we'll compare.

Let's start with the critical points:

• At $x = 0$: $f(0) = (0)^4 + 8(0)^3 + 18(0)^2 + 3$ $f(0) = 0 + 0 + 0 + 3$ $f(0) = 3$

• At $x = -3$: $f(-3) = (-3)^4 + 8(-3)^3 + 18(-3)^2 + 3$ $f(-3) = 81 + 8(-27) + 18(9) + 3$ $f(-3) = 81 - 216 + 162 + 3$ $f(-3) = 246 - 216$ $f(-3) = 30$

Now, let's evaluate the function at the endpoints of the interval $[-4, 1]$:

• At $x = -4$ (the left endpoint): $f(-4) = (-4)^4 + 8(-4)^3 + 18(-4)^2 + 3$ $f(-4) = 256 + 8(-64) + 18(16) + 3$ $f(-4) = 256 - 512 + 288 + 3$ $f(-4) = 547 - 512$ $f(-4) = 35$

• At $x = 1$ (the right endpoint): $f(1) = (1)^4 + 8(1)^3 + 18(1)^2 + 3$ $f(1) = 1 + 8 + 18 + 3$ $f(1) = 30$

So, we have the following function values:

• $f(0) = 3$
• $f(-3) = 30$
• $f(-4) = 35$
• $f(1) = 30$

Keep these numbers handy, because the next step is to compare them all to find our ultimate prize!

Step 3: Determine the Absolute Maximum Value

We've done the heavy lifting, guys! We've found the critical points, checked if they're in our interval, and evaluated the function at these points and at the interval's endpoints. Now comes the moment of truth: identifying the absolute maximum value.

Let's look at the list of function values we calculated:

• $f(0) = 3$
• $f(-3) = 30$
• $f(-4) = 35$
• $f(1) = 30$

Our task is to find the largest number in this list. Let's compare them:

We have $3$, $30$, $35$, and $30$. Clearly, the biggest number among these is $35$. This highest value occurs at $x = -4$.

Therefore, the absolute maximum value of the function $f(x)=x^4+8 x^3+18 x^2+3$ on the closed interval $[-4,1]$ is 35. It’s like finding the highest peak on your specific hiking trail! This value is attained at the left endpoint of our interval. It's a great feeling when everything comes together, right? This process guarantees that we've found the highest possible output for our function within the specified boundaries.

Conclusion

And there you have it, folks! We've successfully navigated the process of finding the absolute maximum value for the function $f(x)=x^4+8 x^3+18 x^2+3$ on the interval $[-4,1]$. By employing the fundamental principles of calculus – finding the derivative, identifying critical points, and evaluating the function at critical points and endpoints – we arrived at our answer. The absolute maximum value is 35, occurring at $x = -4$. This journey through finding absolute extrema is a cornerstone of calculus, demonstrating how derivatives can unlock the secrets of a function's behavior over a given domain. Remember, this method is robust and applies to any continuous function on a closed interval. Keep practicing, and you'll become a pro at spotting these maximums (and minimums!) in no time. The power of calculus is truly amazing when you see it in action like this! Happy calculating, everyone!