2

Tangent line at $x_1$ to polynomial curve $p(x)$ of degree at least $2$ implies $x_1$ is a double root of $p(x) - p^{'}(x_1)(x-x_1)$ ?.

Suppose I have a polynomial function $p(x): \mathbb R \rightarrow \mathbb R$ given by $x \mapsto x^3 + Ax + B$.

Also suppose $L:p^{'}(x_1)(x-x_1)$ is tangent line at the point $x_1$.

Is it always true that $x^3 + Ax + B - p^{'}(x_1)(x-x_1)$ has a double root $x_1$ ?

Is it possible to make this more general ?

Shuzheng
  • 5,533
  • You may want to review your tangent formula. This can be shown by long division. For more general result, try Taylor series of $p$ at $x_1$. – Quang Hoang Jul 28 '14 at 09:01

1 Answers1

1

Yes. Let $f$ be differentiable and let $g(x)=mx+b$ be the equation for the tangent line at $x_0$, that is $g(x_0)=f(x_0)$ and $m = g'(x_0) = f'(x_0)$. Then we consider the function $h(x)=f(x)-g(x)$ and observe that $h(x_0)=f(x_0)-g(x_0)=0$ and $h'(x_0)=f'(x_0)-g'(x_0)=0$ by the very conditions given. Recall that a polynomial (if $h$ is one) has a (at least) $k$fold root at $x_0$ if $h(x_0)=h'(x_0)=\ldots =h^{(k-1)}(x_0)=0$.

Hagen von Eitzen
  • 374,180
  • Thanks very much. Why is it true that the root $x_0$ is $k$ fold if $h(x_0) = h^{'}(x_0) = \dots = h^{(k-1)}(x_0) = 0$ ? – Shuzheng Jul 28 '14 at 09:11
  • @user111854 if $h(x)=(x-x_0)^kh_1(x)$ with some polynomial $h_1$, then $h'(x)=k(x-x_0)^{k-1}h_1(x)+(x-x_0)^kh_1'(x)=(x-x_0)^{k-1}h_2(x)$ with $h_2(x):=kh_1(x)+(x-x_0)h_1'(x)$, i.e. each higher derivative has one factor $(x-x_0)$ less. – Hagen von Eitzen Jul 28 '14 at 09:50