Sorry for my English
Can someone please help me? Taken from Lehmann's book. Group 45, exercise 6. "If a polynomial function $f(x)$, when equal to $0$, has real roots with and even power , each equal to '$a$' ,( like $f(x)= (x+2)(x-1)^2 \to a = 1$) , show that the curve $y=f(x)$ is tangent to the $X$ axis in the point $(a,0)$"
The description justifies that the polynomial function $f(x)$ should have the following form: $$f(x)=p(x)(x-a)^{2k}, \,k\in \mathbb N^*,$$ where $p(x)$ is some other polynomial function. Then it is easy to see that $$f'(x)=p'(x)(x-a)^{2k}+2kp(x)(x-a)^{2k-1}.$$ Since $2k\ge 2$, it follows that $f'(a)=0$, and so the tangent line of the graph of $f(x)$ at $(a,0)$ is simply $$y-0=0(x-a),$$ namely, $$y=0.$$ and so the horizontal axis and the graph of $f(x)$ are tangent to each other at $(a,0)$.
A solution without using calculus
Still we have $$f(x)=p(x)(x-a)^{2k},\,\,k\in\mathbb N^*,$$ where we have further that $p(x)$ does not have the factor $x-a$, and so $p(a)\ne 0$. Without loss of generality, assume that $p(a)>0$. Since $p(x)$ is continuous, there must be some $\varepsilon>0$ such that $p(x)> 0$ on $(a-\varepsilon, a+\varepsilon)$. Thus, we have $f(a)=0$ and $f(x)>0$ on $(a-\varepsilon, a+\varepsilon)\backslash \{a\}$. This also justifies the tangency.
