question_answer

Find a point on the parabola \[y={{(x-3)}^{2}},\] where the tangent is parallel to the chord joining (3, 0) and (4, 1).               

Answer:

Let us apply Lagrange's mean-value theorem for the function \[f(x)={{(x-3)}^{2}}\] in the interval [3, 4]. Now, \[f(x)\] being a polynomial function, it is continuo us on [3, 4]. Also, \[f'(x)=2(x-3),\] which exists for all \[x\in ]3,\,\,4[\] So, \[f(x)\] is differentiable on ] 3, 4[.            Thus, both the conditions of Lagrange's mean-value theorem are satisfied.                          So, there must exist a point \[c\in ]3,\,\,4[\] such that             \[f'(c)=\frac{f(4)-f(3)}{(4-3)}=1\] Now,  \[f'(c)-1\,\,\,\Leftrightarrow \,\,\,2(c-3)=1\,\,\,\Leftrightarrow \,\,\,c=\frac{7}{2}\in ]3,\,\,4[.\] Now, \[x=\frac{7}{2}\]  and \[y={{(x-3)}^{2}}\] \[\Rightarrow \] \[y=\frac{1}{4}.\] Thus, at the point \[\left( \frac{7}{2},\,\,\frac{1}{4} \right)\] on the given curve the tangent is parallel to the chord joining (3, 0) and (4, 1).