Here's a question from my review sheet before midterms. We're learning about interpolation; not sure how to apply it to this.
Find the cubic polynomial f such that f(0) = 0, f(1) = 2, f''(0) = 0, f'(1) = 4.
Thanks.
f(0) = 0, f(1) = 2, f''(0) = 0, f'(1) = 4.
let the polynomial be $\displaystyle ax^3 + bx^2 + cx + d
$
Now
f(0) = 0
so d = 0
so polynomial reduces to $\displaystyle ax^3 + bx^2 + cx
$
f(1) = 2,
a + b + c = 2
$\displaystyle
f'(x) = 3ax^2 + 2bx + c$
$\displaystyle f''(x) = 6ax + 2b$
f''(0) = 0,
so b = 0
f'(1) = 4.
so 3a + 2b = 4
solving we get
a = 4/3
b = 0
c = 2/3
d= 0
so polynomial is $\displaystyle 4/3 x^3 + 2/3 x$
Is there a way to solve it using polynomial interpolation or something? I started out doing it the way you did it but, figured it was wrong since we were supposed to be learning about interpolation. I'll try to solve the other questions using that method for now. Thanks.