Math Problem Statement

Solve these problems

Solve these problems: If the function f(x) = -30 sin(2x) + 15, what are the maximum and minimum values of f(x)? Which of the following statement is true?

Solution

Let's address the two problems in the image step by step.


Problem 21:
The function is given as: f(x)=30sin(2x)+15f(x) = -30 \sin(2x) + 15

We need to find the maximum and minimum values of f(x)f(x).

Solution:

  • The sine function sin(2x)\sin(2x) oscillates between -1 and 1, so the range of sin(2x)\sin(2x) is: 1sin(2x)1-1 \leq \sin(2x) \leq 1

  • Multiply this range by -30: 30×(1)30sin(2x)30×1-30 \times (-1) \leq -30 \sin(2x) \leq -30 \times 1 3030sin(2x)3030 \leq -30 \sin(2x) \leq -30

  • Now add 15 to the entire inequality: 30+1530sin(2x)+1530+1530 + 15 \leq -30 \sin(2x) + 15 \leq -30 + 15 45f(x)1545 \leq f(x) \leq -15

Thus, the maximum value is 4545 and the minimum value is 15-15.

The correct answer is B) max = 45, min = -15.


Problem 22:
We need to identify which of the given statements is true.

Let's examine each option:

A) The period of f(x)=tan(2πx+1)f(x) = \tan(2\pi x + 1) is 12\frac{1}{2}.

  • The period of the tangent function tan(kx)\tan(kx) is πk\frac{\pi}{k}. Here, k=2πk = 2\pi, so the period is: π2π=12\frac{\pi}{2\pi} = \frac{1}{2} This statement is True.

B) The amplitude of f(x)=3tan(2x+3)f(x) = 3 \tan(2x + 3) is 3.

  • The tangent function does not have an amplitude because it has vertical asymptotes and ranges from -\infty to \infty. Therefore, this statement is False.

C) The function f(x)=12sin(2x4)+1f(x) = -\frac{1}{2} \sin(2x - 4) + 1 is shifted vertically downwards.

  • The function f(x)=12sin(2x4)+1f(x) = -\frac{1}{2} \sin(2x - 4) + 1 is shifted vertically upwards by 1 unit, not downwards. Therefore, this statement is False.

D) The function f(x)=3cos(0.5x+3)+2f(x) = 3 \cos(0.5x + 3) + 2 is shifted horizontally to the right.

  • The function f(x)=3cos(0.5x+3)+2f(x) = 3 \cos(0.5x + 3) + 2 has a phase shift. To find the phase shift, solve 0.5x+3=00.5x + 3 = 0: x=60.5=6x = -\frac{6}{0.5} = -6 This indicates a shift to the left by 6 units, not to the right. Therefore, this statement is False.

Thus, the correct answer is A) The period of f(x)=tan(2πx+1)f(x) = \tan(2\pi x + 1) is 12\frac{1}{2}.


Would you like more details or have any other questions? Here are some related questions for practice:

  1. How do you find the period of a sine or cosine function?
  2. What is the range of a general sine function f(x)=Asin(Bx)+Cf(x) = A \sin(Bx) + C?
  3. How does the amplitude of a cosine function affect its graph?
  4. What are the key transformations for the graph of a tangent function?
  5. Can you describe how horizontal phase shifts affect the graph of trigonometric functions?

Tip: Always remember that the amplitude only applies to sine and cosine functions, while tangent and cotangent have no amplitude.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Function Analysis
Periodicity
Amplitude
Phase Shift

Formulas

Range of sin function: -1 ≤ sin(x) ≤ 1
Scaling and shifting formula: y = A sin(Bx + C) + D
Period of tan function: period = π / |B|

Theorems

Sine and Cosine Function Behavior
Periodicity of Trigonometric Functions

Suitable Grade Level

Grades 10-12