Question

Use Rolles Theorem to find a real number in the interval (0,3) Where the derivative of the function f(x)=x(x-3)^2 Vanishes

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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

ROLLE'S THEOREM

Suppose f(x) be a function satisfying three conditions:

1) f(x) is continuous in the closed interval

a ≤ x ≤ b

2) f(x) is differentiable in the open interval

a < x < b

3) f(a) = f(b)

Then there exists at least one point ‘c’ in the open interval (a, b) such that:

f ' (c) = 0

TO DETERMINE

$\sf{For \: the \: function \: \: f(x) =x {(x - 3)}^{2} }$

Use Rolles Theorem to find a real number in the interval (0,3) Where the derivative of the function vanishes

CALCULATION

f(x) is a polynomial.

Therefore

1. f(x) is continuous for all values of x

Therefore f(x) in continuous in the closed interval [ 0, 3 ]

2. f(x) is differentiable for all values of x

Therefore f(x) is differentiable in the open interval ( 0, 3 )

3. f(0) = 0 = f(3)

So f(x) satisfies all three conditions of ROLLE'S THEOREM

Hence by ROLLE'S THEOREM there exists at least one value of x ( say c ) such that

$\sf{{f}{ '}(c) = 0}$

$\implies \sf{ {(c - 3)}^{2} + 2 {c} (c - 3) = 0 \: \: }$

$\implies \sf{ {c}^{2} - 6c + 9 + 2 {c}^{2} - 6c = 0 \: \: }$

$\implies \sf{ 3 {c}^{2} - 12c + 9 = 0 \: \: }$

$\implies \sf{ {c}^{2} - 4c + 3 = 0 \: \: }$

$\implies \sf{ (c - 1)(c - 3) = 0 \: \: }$

$\implies \sf{ c = 1 \: \: or \: \: 3 \: \: }$

$\sf{Clearly \: \: 1 \in \: (0,3)}$

So the value of c is 1

Hence Using Rolles Theorem there exists a real number 1 in the interval (0,3) Where the derivative of the function f(x)=x(x-3)^2 Vanishes

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