# Show that the function F(x)=(x-a)^2 (x-b)^2 +x has the value (a+b)/2 at some point x.

Show that the function $F\left(x\right)=\left(x-a{\right)}^{2}\left(x-b{\right)}^{2}+x$ has the value (a+b)/2 at some point x.
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sweetwisdomgw
It is obvious to say that F(x) is continuous as it is just apolynomial of order 4.
And Now define G(x)=F(x)-(a+b)/2
G(x) is a polynomial of order 4,Henceit is also continuous .
Now find G(a) and G(b),
$⇒G\left(a\right)=F\left(a\right)-\left(a+b\right)/2=\left(a-b\right)/2$
and G(b)=F(b)-(a+b)/2=(b-a)/2
$⇒G\left(a\right)=-G\left(b\right)$
i.e G(a) and G(b) have opposite signs and it is continuoustoo.
Hence the curve G(x) must cut the x-axis at some pointα. i.e G(a)=0
$\therefore G\left(a\right)=F\left(a\right)-\left(a+b\right)/2=0$
$⇒F\left(a\right)=\left(a+b\right)/2$