Aditya Gupta
Last Activity: 6 Years ago
te above ques is really easy.
observe that f’(x)= 2x+1 which is greater than zero for all x in domain of f. so this means that in [0,3] f is an increasing function, which proves that it is an injective function.
now, f(0)= 1 and f(3)= 9+3+1= 13. also, f is a continuous function, so it attains all the values between 1 and 13. hence, we have proved that it is surjective too.
since the function is both injective and surjective, it is a bijection by definition. hence proved
now to find the inverse, simply interchange x and y.
x=y^2+y+1
or y^2+y+1 – x=0
or y= [-1+-root(1 – 4(1-x))]/2
but since y is positive, we drop the negative sign
so, y=f-1(x)=[–1+root(4x – 3))]/2, domain of f-1 being [1,13] and range being [0,3]