Continuity And Radical Expressions: A Mathematical Journey

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Hey guys! Let's dive into a cool math problem that involves proving the continuity of a function. We're gonna be looking at the function h(x) and showing that it's continuous at a specific point. Continuity is a super important concept in calculus, so understanding it is key. This exploration into the mathematical world will strengthen our understanding of functions. Before we jump in, remember that a function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at that point.

So, let's get this show on the road! We're given a function h(x) defined as follows:

h(x) = (√(x+9) - 3) / x if x ≠ 0 and h(0) = 1/6.

Our goal is to show that this function is continuous at x = 0. To do this, we need to prove that the limit of h(x) as x approaches 0 is equal to h(0). Basically, we need to figure out what happens to the function's value as x gets super close to 0, but not quite at 0, and then see if that value matches the function's value at 0. If everything lines up perfectly, we've proven continuity! To find the limit, we can't just plug in x = 0 directly into the first part of the function because that would lead to division by zero, which is a big no-no in math. Instead, we'll need to use some algebraic tricks to simplify the expression.

Let's start by calculating the limit of h(x) as x approaches 0. We'll do this by working with the expression (√(x+9) - 3) / x. The trick here is to get rid of the square root in the numerator. We can do this by multiplying the numerator and denominator by the conjugate of the numerator. The conjugate of √(x+9) - 3 is √(x+9) + 3. Multiplying the numerator and denominator by the conjugate won't change the value of the expression. It's like multiplying by 1.

So we do lim x→0 [(√(x+9) - 3) / x] * [(√(x+9) + 3) / (√(x+9) + 3)]. This simplifies to lim x→0 [(x+9 - 9) / (x * (√(x+9) + 3))]. Further simplification gives us lim x→0 [x / (x * (√(x+9) + 3))]. Now, we can cancel out the x terms in the numerator and denominator, leaving us with lim x→0 [1 / (√(x+9) + 3)].

Now, we can finally plug in x = 0 into this simplified expression! This gives us 1 / (√(0+9) + 3). Simplifying further, we get 1 / (3 + 3), which equals 1/6.

So, we've found that the limit of h(x) as x approaches 0 is 1/6. We are given that h(0) = 1/6. Since the limit of h(x) as x approaches 0 is equal to h(0), we have successfully proven that h(x) is continuous at x = 0. Boom! We used some clever algebra to get around the initial problem of division by zero and were able to find the limit and confirm continuity. Understanding this process really helps in grasping the core ideas behind calculus and understanding how functions behave. This concept is important when analyzing graphs, understanding rates of change, and solving a wide variety of problems in science and engineering. Keep practicing, and these problems will become a breeze!

Working with Radical Expressions: Simplifying the Equation

Alright, let's switch gears and tackle a different type of math problem involving radical expressions. This time, we're going to show that a specific equation involving radicals holds true. We'll be using the rules of exponents and radicals to simplify things and arrive at the desired result. Get ready to dust off those exponent rules and put them to work!

Our task is to prove that √3 * ∛2 * ∜72 = 6 * √¹⁰2. This equation mixes different types of radicals and is designed to test our understanding of how to manipulate them. The goal is to rewrite the left side of the equation so that it looks exactly like the right side. It’s all about using the properties of exponents and radicals to your advantage. This involves converting radicals into fractional exponents, simplifying the numbers under the radicals, and then combining the terms. Before we get into the calculations, let's remember a few key properties. Recall that √a = a^(1/2), ∛a = a^(1/3), and ∜a = a^(1/4). Also, when you multiply terms with the same base, you add their exponents. And when you have a power raised to another power, you multiply the exponents.

Let's start by rewriting the left side of the equation using fractional exponents. We can rewrite √3 as 3^(1/2), ∛2 as 2^(1/3), and ∜72 as 72^(1/4).

So, our equation becomes 3^(1/2) * 2^(1/3) * 72^(1/4). Now, we can break down the number 72 into its prime factors. 72 can be expressed as 2³ * 3². So we can rewrite 72^(1/4) as (2³ * 3²)^(1/4).

Using the power of a product rule, we can distribute the exponent 1/4 to both 2³ and 3². This gives us 2^(3/4) * 3^(2/4), which simplifies to 2^(3/4) * 3^(1/2).

Now, let's substitute these back into our original equation. We now have 3^(1/2) * 2^(1/3) * 2^(3/4) * 3^(1/2). We can combine the terms with the same base. The 3^(1/2) terms can be combined: 3^(1/2) * 3^(1/2) = 3^(1/2 + 1/2) = 3¹ = 3. The terms with base 2 are 2^(1/3) and 2^(3/4). They combine to give us 2^(1/3 + 3/4). Adding the exponents, we get 2^(4/12 + 9/12) = 2^(13/12).

So now our equation is 3 * 2^(13/12). We still need to arrive at 6 * √¹⁰2, so let's take a closer look at 2^(13/12). We can rewrite the exponent 13/12 as 1 + 1/12. Which means, we can rewrite 2^(13/12) as 2¹ * 2^(1/12). The equation is now 3 * 2 * 2^(1/12) which simplifies to 6 * 2^(1/12).

To get our target form, let's think about how we can get a tenth root instead of a twelfth root. We can express the root 1/12 into tenth root by doing some slight adjustment to the power. It is impossible to get the final answer with the above method, so let's try the simpler way.

Let's try a different approach.

We are trying to prove: √3 * ∛2 * ∜72 = 6 * √¹⁰2. Let's rewrite the left side √3 * ∛2 * ∜72 using exponents: 3^(1/2) * 2^(1/3) * 72^(1/4). Decompose 72 into prime factors: 72 = 2^3 * 3^2. Then, the equation becomes: 3^(1/2) * 2^(1/3) * (2^3 * 3^2)^(1/4). Applying the power of a product, we have: 3^(1/2) * 2^(1/3) * 2^(3/4) * 3^(2/4). Simplify: 3^(1/2) * 2^(1/3) * 2^(3/4) * 3^(1/2). Combine like terms. 3^(1/2) * 3^(1/2) = 3^(1) = 3. 2^(1/3) * 2^(3/4) = 2^(1/3 + 3/4) = 2^(13/12). So, the equation becomes: 3 * 2^(13/12). Now, let's change this expression to the form of 6 * √¹⁰2. Let's start with 6. We know that 6 = 2 * 3. Then our previous equation becomes: 2 * 3 * 2^(13/12) = 2^(1) * 3^(1) * 2^(13/12). Let's change the equation 6 * √¹⁰2 = 6 * 2^(1/10). The left side of the original equation can not simply to the right side by simply rearranging the terms and combining the powers. There might be a typo in the original equation. Let's assume the target equation is √3 * ∛2 * ∜72 = 6 * √¹²2. If that's the case, we can get: 3 * 2^(13/12) = 3 * 2 * 2^(1/12) = 6 * 2^(1/12) = 6 * √¹²2. If the original equation has a typo, then we have successfully demonstrated how to solve the problem. This involves transforming the radicals into fractional exponents, simplifying the numbers, and using properties of exponents to combine like terms. The key is to break down the expressions into their simplest components and then use the rules to combine them.

Conclusion: Mastering Mathematical Concepts

Alright guys, we've made it through two interesting math problems! We've proven the continuity of a function and played around with radical expressions, and used some cool math tricks along the way. We've seen how important it is to have a solid grasp of the fundamentals, like limits, conjugates, and exponent rules. Keep practicing and challenging yourself with new problems, and you'll be amazed at how your skills improve. Mathematics is a fascinating journey, and every problem you solve is a step forward. So, keep exploring, keep learning, and most importantly, have fun with it! Keep in mind, with consistent effort and a positive attitude, you can absolutely conquer these concepts. Until next time, happy calculating!