Let # f(x) = a x^3 + b x^2 + c x + d #.
Firstly, note that the leading coefficient is 1, which means # a = 1 #. So # f(x) = x^3 + b x^2 + c x + d #.
Now, calculate the derivatives of # f #. We get # f'(x) = 3 x^2 + 2 b x + c #, and # f''(x) = 6 x + 2 b #.
Secondly, since there is an inflection point at # x = 1 #, we know that # f''(1) = 0 #, which implies # 6 + 2b = 0 #, or # b = -3 #. So # f(x) = x^3 - 3 x^2 + c x + d #, and # f'(x) = 3 x^2 - 6 x + c #.
Thridly, since there is an critical point at # x = 2 #, we know that # f'(2) = 0 #, which implies # 3 xx 2^2 - 6 xx 2 + c = 0 #, or # c = 0 #. So # f(x) = x^3 - 3 x^2 + d #.
Finally, we use the fact that the curve passes through # (1, 7) #, so # f(1) = 7 #. Substituting, we get # 1 - 3 + d = 7 #, so # d = 9 #. So # f(x) = x^3 - 3 x^2 + 9 #.
