Answer:
a) [tex]f^{-1} (x) = \frac{1}{2} Cos^{-1} (\frac{x-1}{3} ) +\frac{3}{2}[/tex]
The inverse of given function
[tex]f^{-1} (x) = \frac{1}{2} Cos^{-1} (\frac{x-1}{3} ) +\frac{3}{2}[/tex]
Step-by-step explanation:
Step(i):-
Given function f(x) = 3 cos (2 x -3) + 1
Let y = f(x) = 3 cos (2 x -3) + 1
y = 3 cos (2 x -3) + 1
⇒ y - 1 = 3 cos (2 x -3)
⇒ [tex]cos ( 2 x - 3 ) =\frac{y -1}{3}[/tex]
⇒[tex]cos ^{-1} ( cos (2 x - 3)) = Cos^{-1} (\frac{y-1}{3} )[/tex]
We know that inverse trigonometric equations
cos⁻¹(cosθ) = θ
[tex]2 x - 3 = Cos^{-1} (\frac{y-1}{3} )[/tex]
[tex]2 x = Cos^{-1} (\frac{y-1}{3} ) +3[/tex]
[tex]x = \frac{1}{2} Cos^{-1} (\frac{y-1}{3} ) +\frac{3}{2}[/tex]
Step(ii):-
we know that y= f(x)
The inverse of the given function
[tex]x = f^{-1} (y)[/tex]
[tex]f^{-1} (y) = \frac{1}{2} Cos^{-1} (\frac{y-1}{3} ) +\frac{3}{2}[/tex]
The inverse of given function in terms of 'x'
[tex]f^{-1} (x) = \frac{1}{2} Cos^{-1} (\frac{x-1}{3} ) +\frac{3}{2}[/tex]
conclusion:-
The inverse of given function
[tex]f^{-1} (x) = \frac{1}{2} Cos^{-1} (\frac{x-1}{3} ) +\frac{3}{2}[/tex]
