Question #311456

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).


1
Expert's answer
2022-03-17T18:39:11-0400

(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)

f1(x)f^{-1}(x)

f(x)=2cos(x3)f(x)=2cos(x-3)

y=2cos(x3)y=2cos(x-3)

Interchanging x and y,

x=2cos(y2)x=2cos(y-2)

Solve for y

x2=cos(y2)\frac{x}{2}=cos(y-2)

y2=cos1(x2)y-2=cos^{-1}(\frac{x}{2})

y=cos1(x2)+2y=cos^{-1}(\frac{x}{2})+2


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