if $f(x)=x^3+ax^2+bx$,show that $g'(x_{1})=0$

Question

for any real $a,b$, Assmue $x_{1}<x_{2}$ such $f'(x_{1})=f'(x_{2})=0$.where $f(x)=x^3+ax^2+bx$ and let $x_{1}+2x_{0}=3x_{2}$,$g(x)=f(x)-f(x_{0})$. show that $$g(x)=0$$ only two real roots?

since $f'(x)=3x^2+2ax+b$,so we have $$x_{1}+x_{2}=-\dfrac{2a}{3},~~~~~x_{1}x_{2}=\dfrac{b}{3}$$ and $$g(x)=f(x)-f(x_{0})=(x-x_{0})(x^2+(x_{0}+a)x+x^2_{0}+ax_{0}+b)$$ then fell it ugly to

Answer 1

If $g(x) = f(x) - f(x_0)$, then by differentiating, we yield :

$$g'(x) = \big(f(x)-f(x_0)\big)' =f'(x) - \big(f(x_0)\big)' \Rightarrow g'(x) = f'(x)$$

if $x_0 \in \mathbb R$ is a constant, which would mean that $f(x_0)$ is a constant as well.

Thus, to show that $g'(x_1) =0$, all you have to do is show that $f'(x_1) = 0$, which is true by the hypothesis.

Answer 2

Note that $x_0>x_2$. It suffices to show that $f(x_1)=f(x_0)$. This equality holds iff $x_1^3+ax_1^2+bx_1=x_0^3+ax_0^2+bx_0$ holds iff $x_1^2+x_1x_0+x_0^2+a(x_1+x_0)+b=0$ holds. By the relation of roots and coefficients , one has $3x_1+3x_2=-2a, 3x_1x_2=b$ and use $3x_2=x_1+2x_0$ to eliminate $3x_2$. You can get $a=-2x_1-x_0, b=x_1^2+2x_1x_0$. Using this two equalities to eliminate variables $a$ and $b$, easy to check the above equality holds, we are done.