Please answer part C of this homework practice question. Thank you. For Part A, I and my friends got D: (inf,-1]U[1,inf]; R: [0,pi/2)U[pi,3pi/2). For Part B, we got f^(-1) (sec5) =2pi-5; sec (f(-1)(5)) = 5. I hope this helps in answering Part C. 1. We restrict the function secz on [0, π/2) U[, 3/2) to get an one-to-one function. We call it f(2). Wedefine the inverse of this function to be f-¹(r).(a) Find the domain and range of f-¹. No need to justify your answer to this part.Final AnswerFinal AnswerDomain:Range:(b) Compute f-¹(sec 5) and sec(f-¹(5)). No need to justify your answer to this part.Final AnswerFinal Answerƒ-¹(sec 5) =sec(ƒ-¹(5)) =(c) Sketch the graph of secof-1. Make sure to label your axes and that you have the correct domain.

1. We restrict the function secz on [0, π/2) U[,…

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Please answer part C of this homework practice question. Thank you. For Part A, I and my friends got D: (inf,-1]U[1,inf]; R: [0,pi/2)U[pi,3pi/2). For Part B, we got f^(-1) (sec5) =2pi-5; sec (f(-1)(5)) = 5. I hope this helps in answering Part C.