Mean Value Theorem ( Lagrange's), How to take Domain in this type of questions?)

Question

Find a point on the graph of y=x3(cube)  where the tangent is parallel to the chord joining (1,1) and (3,27).

Answer 1

$f(x) = x^3 \ in \ the \ interval \ [1 , 3]$               $...........................(1)$

1. $f(x)$ being polynomial is continuous in $[1,3]$

2. $f(x)$ being polynomial is differentiable in $[1,3]$

Both are satisfying to apply Lagrange's mean value theorem. According to this there exists a point $c$ in $[1,3]$ such that

$f'(c) = \underline{f(b)-f(a)}$

                       $b - a$

$f(3) = 3^3 = 27$

$f(1) = 1^3 = 1$

differentiating w.r.t to $x$, we get

$f'(x) = 3x^2$

$f'(c) = 3c^2$

$3c^2 = (27 - 1) / (3 - 1)$

$3c^2 = 26 / 2$

$c = \pm \sqrt{13*3/3*3}$

$c = \pm \sqrt{13*3}/3$

$c = \pm \sqrt{39}/3$

As $c \ \varepsilon [1 , 3]$

So $c = \sqrt{39}/3$

 $x = \sqrt{39}/3$

From $(1) , y = (13\sqrt{39})/3$

$\therefore \bigl(\begin{smallmatrix} \sqrt{39}/3, & (13\sqrt{39})/3 \end{smallmatrix}\bigr)$ is the point on the given curve

$f(x) =x^3$, where tangent is parallel to the chord joining $(1,1)$ and $(3,27)$

Comments

that was clear.

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The y coordinate of the point should be $13(\sqrt{39}) / 9$