Let  f (x)  =  sin(  
πx
8
  ) + 4x2 + 8,  x  ≥  1. Find  f −1(264).


We are tasked with finding \( f^{-1}(264) \), which means we want to determine the value of \( x \) such that \( f(x) = 264 \). Given the function \( f(x) = \sin\left(\frac{\pi x}{8}\right) + 4x^2 + 8 \), we start by setting it equal to 264:

\[
\sin\left(\frac{\pi x}{8}\right) + 4x^2 + 8 = 264.
\]

### Step 1: Isolate the quadratic term
Subtract 8 from both sides:

\[
\sin\left(\frac{\pi x}{8}\right) + 4x^2 = 256.
\]

### Step 2: Eliminate the sine term (approximation method)
For simplicity, let's approximate \( \sin\left(\frac{\pi x}{8}\right) \), because it's typically small compared to \( 4x^2 \) as \( x \) grows larger. So, we'll begin by ignoring the sine term:

\[
4x^2 \approx 256.
\]

Solve for \( x^2 \):

\[
x^2 = \frac{256}{4} = 64.
\]

Thus, \( x = \sqrt{64} = 8 \).

### Step 3: Check and refine the approximation
To check the accuracy, substitute \( x = 8 \) back into the original function:

\[
f(8) = \sin\left(\frac{\pi \cdot 8}{8}\right) + 4(8)^2 + 8 = \sin(\pi) + 4(64) + 8 = 0 + 256 + 8 = 264.
\]

Since \( f(8) = 264 \), we confirm that \( x = 8 \) is correct.

Thus, \( f^{-1}(264) = 8 \).

Would you like further details on this, or have any questions?

Here are five related questions for further practice:
1. Find \( f^{-1}(300) \) for the same function.
2. Solve \( f(x) = 400 \) for \( x \).
3. Given \( g(x) = \cos\left(\frac{\pi x}{8}\right) + 4x^2 + 8 \), find \( g^{-1}(300) \).
4. What is the range of \( f(x) \) for \( x \geq 1 \)?
5. How would the result change if \( f(x) = \sin\left(\frac{\pi x}{6}\right) + 4x^2 + 8 \)?

**Tip:** When solving inverse problems, it’s useful to check if the trigonometric part contributes significantly or can be approximated.

Let f(x) = sin(πx/8) + 4x^2 + 8, x ≥ 1. Find f−1(264).

Learn how to find the inverse of the function f(x) = sin(πx/8) + 4x^2 + 8, where f−1(264) is solved step-by-step using algebraic manipulation and trigonometric approximation techniques. This problem is suited for high school students in Grades 10-12, covering inverse functions and quadratics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation