Math Problem Statement

limπ‘₯β†’11.5(4π‘₯2+8π‘₯βˆ’6)3/2

Solution

The expression given is:

lim⁑xβ†’11.5(4x2+8xβˆ’6)32\lim_{x \to 11.5} \left( 4x^2 + 8x - 6 \right)^{\frac{3}{2}}

To evaluate this limit, we can follow these steps:

  1. Substitute x=11.5x = 11.5 directly into the expression 4x2+8xβˆ’64x^2 + 8x - 6.
  2. Compute the expression inside the parentheses.
  3. Raise the result to the power of 32\frac{3}{2}.

Let's calculate it step by step.

Step 1: Substitute x=11.5x = 11.5 into the expression

4(11.5)2+8(11.5)βˆ’64(11.5)^2 + 8(11.5) - 6

Step 2: Calculate the expression inside the parentheses

4(132.25)+92βˆ’6=529+92βˆ’6=6154(132.25) + 92 - 6 = 529 + 92 - 6 = 615

Step 3: Raise the result to the power of 32\frac{3}{2}

61532=(61512)3=(615)3615^{\frac{3}{2}} = (615^{\frac{1}{2}})^3 = (\sqrt{615})^3

The square root of 615 is approximately 24.8, so:

(24.8)3β‰ˆ15252.35(24.8)^3 \approx 15252.35

Final Answer:

lim⁑xβ†’11.5(4x2+8xβˆ’6)32=15252.35\lim_{x \to 11.5} \left( 4x^2 + 8x - 6 \right)^{\frac{3}{2}} = 15252.35

Would you like a more detailed explanation or have any further questions?

Here are five related questions for further exploration:

  1. How do we compute limits involving polynomials?
  2. What is the significance of the exponent 32\frac{3}{2} in this problem?
  3. How do square roots affect the calculation of limits?
  4. Can limits be evaluated for undefined expressions?
  5. How do you handle limits that approach infinity?

Tip: When evaluating limits, direct substitution is often the first and simplest method to try, especially if the function is continuous at the point of interest.

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Math Problem Analysis

Mathematical Concepts

Limits
Polynomials
Exponentiation

Formulas

Limit definition
Exponentiation rules

Theorems

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Suitable Grade Level

Grades 11-12