## University Calculus: Early Transcendentals (3rd Edition)

$F(x)$ does take on the value $(a+b)/2$ for some value of $x$. The proof is discussed in detail below.
$$F(x)=(x-a)^2(x-b)^2+x$$ To show that $F(x)=\frac{a+b}{2}$ for some value of $x$, we need to find 2 values $x_1$ and $x_2$ so that $F(x_1)\lt \frac{a+b}{2}\lt F(x_2)$ - Take $x=a$: $F(a)=(a-a)^2(a-b)^2+a=0\times(a-b)^2+a=a$ - Take $x=b$: $F(b)=(b-a)^2(b-b)^2+b=(b-a)^2\times0+b=b$ Here I would take an arbitrary assumption that $a\lt b$. You can also take the other way around. Since $a\lt b$, we have $$a+a\lt a+b$$ $$\frac{2a}{2}\lt\frac{a+b}{2}$$ $$a\lt\frac{a+b}{2}$$ Similary, since $a\lt b$, we have $$a+b\lt b+b$$ $$\frac{a+b}{2}\lt\frac{2b}{2}$$ $$\frac{a+b}{2}\lt b$$ Therefore, $$a\lt\frac{a+b}{2}\lt b$$ $$F(a)\lt\frac{a+b}{2}\lt F(b)$$ - On $[a,b]$, we have: $\lim_{x\to c}F(x)=(c-a)^2(c-b)^2+c=F(c)$. So $F(x)$ is continuous on $[a,b]$. Therefore, according to the Intermediate Value Theorem, there exists a value of $x=c\in[a,b]$ such that $F(c)=(a+b)/2$