Determine whether each of these functions is a bijection from R toR.a) f(x) = -3x + 4b) f(x) = -3x2 +7c) f(x) = (x+1)/(x+2)d) f(x) = x5 +1"

Answer:a) and d) are bijections. b) and c) are notStep-by-step explanation:a) Every linear non constant function is a bijection. We can easily find the inverse of f by making a simple calculus. If y is on the function image, we have y = -3x + 4 for certain x, theny- 4 = -3x-(y-4)/3 = xtherefore [tex]f^{-1}(x) = -(x-4)/3 [/tex]b) -3 * X² + 4  is not a bijection because quadratic funtions arent bijective. If you evaluate in opposite values you will obtain the same result. For example f(-1) = f(1) = 6c) (x+1)/(x+2) is not a bijection. It isnt even defined in -2 because the denominator is equal to 0 if X= -2 and we cant divide by 0. A bijective function from R to R must be defined in every element of R. In general, homographic non linear functions are not bijective for the same reason this function is not.d) [tex] X^5+1 [/tex] is bijective. There isnt a simple argument we can use to conclude this. We have no other choice than trying to find the inverse function by making a calculus.Y = [tex] X^5+1 [/tex]Y-1 = [tex] X^5[/tex][tex] (Y-1)^{1/5} = X [/tex]Not that since 5 is odd, we can calculate [tex] (Y-1)^{1/5} [/tex] independently of which value Y-1 takes. Therefore [tex] f^{-1}(x) = (x-1)^{1/5} [/tex] , and we can conclude that f is bijective.I hope this helps you!