Math Problem Statement

give me the answer

Find the range of the function f(x) = log2((sin(x) - cos(x) + 3/2) / sqrt(2))

Solution

The function provided is:

f(x)=log2(sinxcosx+322)f(x) = \log_2\left(\frac{\sin x - \cos x + \frac{3}{2}}{\sqrt{2}}\right)

To find the range of this function, we need to determine the range of the expression inside the logarithm and ensure it is positive (since the logarithm is only defined for positive numbers).

Step-by-step Solution:

  1. Simplify the Inside Expression:

    Let's consider the expression inside the logarithm:

    y=sinxcosx+322y = \frac{\sin x - \cos x + \frac{3}{2}}{\sqrt{2}}

  2. Find the Range of sinxcosx\sin x - \cos x:

    Recall that the general form of sinxcosx\sin x - \cos x can be rewritten using a phase shift:

    sinxcosx=2sin(xπ4)\sin x - \cos x = \sqrt{2} \sin\left(x - \frac{\pi}{4}\right)

    The range of sin(xπ4)\sin\left(x - \frac{\pi}{4}\right) is [1,1][-1, 1], so the range of 2sin(xπ4)\sqrt{2} \sin\left(x - \frac{\pi}{4}\right) is [2,2][- \sqrt{2}, \sqrt{2}].

  3. Adjust the Range with the Constant:

    Now add 32\frac{3}{2} to the range [2,2][- \sqrt{2}, \sqrt{2}]:

    [2,2]+32=(322,32+2)[-\sqrt{2}, \sqrt{2}] + \frac{3}{2} = \left(\frac{3}{2} - \sqrt{2}, \frac{3}{2} + \sqrt{2}\right)

    The approximate range becomes:

    (321.414,32+1.414)=(0.086,2.914)\left(\frac{3}{2} - 1.414, \frac{3}{2} + 1.414\right) = (0.086, 2.914)

  4. Divide by 2\sqrt{2}:

    Now, divide this range by 2\sqrt{2}:

    (0.0862,2.9142)(0.061,2.06)\left(\frac{0.086}{\sqrt{2}}, \frac{2.914}{\sqrt{2}}\right) \approx (0.061, 2.06)

  5. Determine the Range of f(x)f(x):

    Since f(x)=log2(y)f(x) = \log_2(y), where yy must be positive:

    y(0.061,2.06)y \in (0.061, 2.06)

    Therefore, the range of f(x)f(x) is:

    (log2(0.061),log2(2.06))\left(\log_2(0.061), \log_2(2.06)\right)

    Calculating these logarithms:

    log2(0.061)4.03,log2(2.06)1.05\log_2(0.061) \approx -4.03, \quad \log_2(2.06) \approx 1.05

    Thus, the range of f(x)f(x) is approximately:

    (4.03,1.05)(-4.03, 1.05)

Conclusion:

The range of the function f(x)=log2(sinxcosx+322)f(x) = \log_2\left(\frac{\sin x - \cos x + \frac{3}{2}}{\sqrt{2}}\right) is approximately (4.03,1.05)(-4.03, 1.05).

Would you like more details on any part of the solution or have any questions?

Related Questions:

  1. What is the domain of the function f(x)=log2(x)f(x) = \log_2(x)?
  2. How do you transform trigonometric expressions into simpler forms?
  3. What are the properties of logarithmic functions?
  4. How does a phase shift affect the range of a sine or cosine function?
  5. What is the significance of the base in logarithmic functions?

Tip:

When dealing with logarithmic functions, always ensure the argument is positive, as logarithms are only defined for positive real numbers.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Trigonometry
Range of Functions

Formulas

f(x) = log2(y)
y = (sin(x) - cos(x) + 3/2) / sqrt(2)
sin(x) - cos(x) = sqrt(2) * sin(x - π/4)

Theorems

Logarithmic function is defined for positive numbers only
Range of trigonometric functions

Suitable Grade Level

Grades 11-12 (Pre-Calculus/Calculus level)