Since [tex]r(x) = ax^3+bx^2+cx+d[/tex], we have first and second derivatives
[tex]r'(x) = 3ax^2 + 2bx + c[/tex]
[tex]r''(x) = 6ax + 2b[/tex]
[tex]r(x)[/tex] is supposed to pass through the points C (2, 2) and D (3, 2), so
[tex]r(2) = 8a + 4b + 2c + d = 2[/tex]
[tex]r(3) = 27a + 9b + 3c + d = 2[/tex]
[tex]\implies (27a+9b+3c+d)-(8a+4b+2c+d)=2-2[/tex]
[tex]\implies \boxed{19a + 5b + c = 0}[/tex]
"agreement with [tex]q(x)[/tex]" entails having the same first and second derivatives as [tex]q[/tex] at the point C :
[tex]q(x) = 5x-2x^2 \implies q'(x) = 5-4x \implies q'(2) = -3[/tex]
[tex]\implies r'(2) = \boxed{12a + 4b + c = -3}[/tex]
[tex]q'(x) = 5-4x \implies q''(x) = -4 \implies q''(2) = -4[/tex]
[tex]\implies r''(2) = \boxed{12a + 2b = -4}[/tex]
Solve the indicated equations for a, b, and c, and subsequently for d :
[tex](12a + 4b + c) - (19a + 5b + c) = -3 - 0 \implies -7a - b = -3 \implies 7a + b = 3[/tex]
[tex](12a + 2b) - 2 (7a + b) = -4 - 2\times3 \implies -2a = -10 \implies \boxed{a=5}[/tex]
[tex]12a+2b=-4 \implies 2b = -64 \implies \boxed{b = -32}[/tex]
[tex]19a + 5b + c = 0 \implies \boxed{c = 65}[/tex]
[tex]8a + 4b + 2c + d = 2 \implies \boxed{d = -40}[/tex]
Then the cubic joining C and D is given by
[tex]r(x) = \boxed{5x^3 - 32x^2 + 65x - 40}[/tex]
