I've also attached the answers but dont understand , could you pls explain accordingly? thanks

(c) A continuous function g : [0, co) > R satises for all 1' E [0, co) the inequalities :r: + 1 g(:1:1). (iii) Hence explain why 9 has no global maximum on [0, oo). [9 marks] [This was the most challenging part of the paper, as it required improvisation when approaching the proof and accuracy when justifying answers. Most students struggled with this proof and received 2, 3 or 4 of the available 9 marks] For part (0,1 CAO mark for choosing a reference point r E [0, oc) (r : 0 is the simplest choice) and splitting the domain [0, 00) of 5: based on that choice; i.e., for expressing [0. 00) as [(1.30) 2 [(),21' + 2] U (2r + 2, 00). Then, 1 CAO mark for introducing a global minimum of g on [0. 2r + 2]; Le, a point .12\