Answer to Question #311456 in Trigonometry for bookaddict

Question

The function 𝑓(𝑥)=2𝑐𝑜𝑠 𝑥−3 is defined for the domain 0≤𝑥≤𝜋/2

a. Find the range of 𝑓(𝑥)

b. Find 𝑓^−1(𝑥).

Answer 1

(i)

F(x)=2cos(x-3)

We begin by finding the magnitude of the trig term 

2cos(x-3)

 by taking the absolute value of the coefficient.

=2

The lower bound of the range for cosine is found by substituting the negative magnitude of the coefficient into the equation.

y=-2

The upper bound of the range for cosine is found by substituting the positive magnitude of the coefficient into the equation.

y=2

The range is 

-2≤y≤2

Interval Notation:

[-2,2]

Set-Builder Notation:

{y|−2≤y≤2}

Hence the range becomes,

Range: [−2,2], {y|−2≤y≤2}

(ii)

$f^{-1}(x)$

$f(x)=2cos(x-3)$

$y=2cos(x-3)$

Interchanging x and y,

$x=2cos(y-2)$

Solve for y

$\frac{x}{2}=cos(y-2)$

$y-2=cos^{-1}(\frac{x}{2})$

$$y=cos^{-1}(\frac{x}{2})+2$$