Does there exist a degree-$4$ polynomial such that $f(1)=0$, $f(0)=1$, $f '(1)=0$ and $f'(0)=1$?

Question

This is a problem in abstract algebra about polynomials that I don't know how to answer. I know that $1$ is a a double root, but I can't get much more from it. Thank you for reading.

Answer 1

Since $1$ is double root for $f$ we have $$f(x) = (x-1)^2(ax^2+bx+c)$$

Since $f(0) =1$ we get $c=1$ and since $f'(0)=1 $ we get $b=3$. So we have a family of polynomial which satisfies condition.

Answer 2

HINT: why don't you just write down a general polynomial of degree 4 and apply all the conditions required by the problem?

Answer 3

assuming $$f(x)$$ is given by $$f(x)=ax^4+bx^3+cx^2+dx+e$$ then $$f(1)=a+b+c+d+e=0$$ $$f(0)=e=1$$ with $$f'(x)=4ax^3+3bx^2+2cx+d$$ we get $$f'(1)=4a+3b+2c+d=0$$ $$f'(0)=d=1$$

Answer 4

First $f(x)=ax^4+bx^3+cx^2+dx+e$... Now $a+b+c+d+e=0$, $e=1$, $4a+3b+2c+d=0$ and $d=1$... So $a+b+c=-2$. So $c=-2-a-b$. So $4a+3b+2(-2-a-b)+1=0$. So $2a+b=3$. So $c=-5+a$. So we get $ax^4+(3-2a)x^3+(-5+a)x^2+x+1$ for our polynomial. $a$ can be anything... other than zero...