Question

Please help me with the below question.

Answer 1

Since
`r(x) = ax^3+bx^2+cx+d`, we have first and second derivatives

`r'(x) = 3ax^2 + 2bx + c`

`r''(x) = 6ax + 2b`

`r(x)` is supposed to pass through the points C (2, 2) and D (3, 2), so

`r(2) = 8a + 4b + 2c + d = 2`

`r(3) = 27a + 9b + 3c + d = 2`

`\implies (27a+9b+3c+d)-(8a+4b+2c+d)=2-2`

`\implies \boxed{19a + 5b + c = 0}`

"agreement with
`q(x)`" entails having the same first and second derivatives as
`q` at the point C :

`q(x) = 5x-2x^2 \implies q'(x) = 5-4x \implies q'(2) = -3`

`\implies r'(2) = \boxed{12a + 4b + c = -3}`

`q'(x) = 5-4x \implies q''(x) = -4 \implies q''(2) = -4`

`\implies r''(2) = \boxed{12a + 2b = -4}`

Solve the indicated equations for a, b, and c, and subsequently for d :

`(12a + 4b + c) - (19a + 5b + c) = -3 - 0 \implies -7a - b = -3 \implies 7a + b = 3`

`(12a + 2b) - 2 (7a + b) = -4 - 2*3 \implies -2a = -10 \implies \boxed{a=5}`

`12a+2b=-4 \implies 2b = -64 \implies \boxed{b = -32}`

`19a + 5b + c = 0 \implies \boxed{c = 65}`

`8a + 4b + 2c + d = 2 \implies \boxed{d = -40}`

Then the cubic joining C and D is given by

`r(x) = \boxed{5x^3 - 32x^2 + 65x - 40}`