Find a point on the parabola $y=(x-4)^{2}$, where the tangent is parallel to the chord joining $(4,0)$ and $(5,1)$.
Let:
$f(x)=(x-4)^{2}=x^{2}-8 x+16$
The tangent to the curve is parallel to the chord joining the points $(4,0)$ and $(5,1)$.
Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$.
$\therefore a=4, b=5$
The polynomial function is everywhere continuous and differentiable.
So, $x^{2}-8 x+16$ is continuous on $[4,5]$ and differentiable on $(4,5)$.
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists $c \in(4,5)$ such that $f^{\prime}(c)=\frac{f(5)-f(4)}{5-4}$.
Now,
$f(x)=x^{2}-8 x+16 \Rightarrow f^{\prime}(x)=2 x-8, f(5)=1, f(4)=0$
$\therefore f^{\prime}(x)=\frac{f(5)-f(4)}{5-4} \Rightarrow 2 x-8=\frac{1}{1} \Rightarrow 2 x=9 \Rightarrow x=\frac{9}{2}$
Thus, $c=\frac{9}{2} \in(4,5)$ such that $f^{\prime}(c)=\frac{f(5)-f(4)}{5-4}$
Clearly,
$f(c)=\left(\frac{9}{2}-4\right)^{2}=\frac{1}{4}$
Thus, $(c, f(c))$, i.e. $\left(\frac{9}{2}, \frac{1}{4}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points $(4,0)$ and $(5,1)$
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