MHB Karla's question at Yahoo Answers (Intermediate Value Theorem).

[Fernando Revilla]

Here is the question:

Show that the function f(x)=(x-a)^2*(x-b)^2+x takes on the value (a+b)/2 for some value of x.?

Here is a link to the question:

Show that the function f(x)=(x-a)^2*(x-b)^2+x takes on the value (a+b)/2 for some value of x.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Karla,

The function $f(x)=(x-a)^2(x-b)^2+x$ is continuos in $\mathbb{R}$ (polynomical function). Besides, $f(a)=a$ and $f(b)=b$.

But $\dfrac{a+b}{2}$ is the middle point of the segment with endpoints $a$ and $b$ (no matter if $a<b$, $b<a$ or $a=b$) so, $\dfrac{a+b}{2}$ is included between $a$ and $b$. According to the Intermediate Value Theorem, there exists $x\in\mathbb{R}$ such that $f(x)=\dfrac{a+b}{2}$.