by Guest » Sat May 09, 2020 3:09 pm
The first problem is
"Find the cubic polynomial f(x) such that f(2)= -1, f'(2)= -24, f''(2)= -26, and f'''(2)= -12".
It would have helped if you had told use what it was that was stopping you from solving tbe problem!
Do you know what a "cubic polynomial" is? Any cubic polynomial can be written in the form [tex]f(x)= ax^3+ bx^2+ cx+ d[/tex] for some number a, b, c, d. Finding f means finding those numbers. There are four so we need four equations or other conditions. And "f(2)= -1, f'(2)= -24, f''(2)= -26, and f'''(2)= -12" are precisely four conditions!
Do you know what f', f'', f''', and f'''' mean? Those are the first, second, third, and fourth derivatives:
f(2)= a(2^3)+ b(2^2)+ c(2)+ d= 8a+ 4b+ 2c+ d= -1.
f'(x)= 3ax^2+ 2bx+ c so
f'(2)= 3a(2^2+ 2b(2)+ c= 12a+ 4b+ c= -24.
f''(x)= 6ax+ 2b so
f''(2)= 12a+ 2b= -26
f'''(x)= 6a so
f'''(2)= 6a= -12.
Solve the four equations 6a= -12, 12a+ 2b= -26, 12a+ 4b+ c= -24, and 8a+ 4b+ 2c+ d= -1.
That is a "triangular system" so it is particularly easy to solve.
Solve 6a= -12 for a, put that value of a into 12a+ 2b= -26 and solve for b, put those values of a and b into 23a+ 4b+ c= -24 and solve for c, put those values of a, b, and c into 8a+ 4b+ 2c+ d= -1 and solve for d.