Question

(b) Apply Rolle's Theorem to find position of real zeros of f'(x)
where f(x) = x (x - 1)(x-2)(x-3).​

Answer 1

The real zeroes for f'(x) are

x1 = 0.38197, x2 = 1.5 and x3 = 2.61803.

Given:

A Function f(x) = x (x - 1)(x-2)(x-3).

To Find:

The position of real zeros of f'(x) using Rolle's Theorem.

Solution:

• Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b. Here, we do not have any intervals.

f(x) = x (x - 1)(x-2)(x-3)

f(x) = x( x-1 ) ( x^2 - 5x + 6)

f(x) = x ( x^3 - 5x^2 + 6x - x^2 + 5x - 6 )

f(x) = x^4 - 6x^3 + 11x^2 - 6x

Now to find f'(x) we have to differentiate the above equation with respect to 'x' on both sides.

We get,

f'(x) = 4*x^(4-1) - 6*3*x^(3-1) + 11*2*x^(2-1) - 6(1)

f'(x) = 4x^3 - 18x^2 + 22x - 6

Now we need to apply Rolle's Theorem to find the position of real zeros of f'(x).

f'(x)

= 4x^3 - 18x^2 + 22x - 6

=  2 * (2x^3 - 9x^2 + 11x - 3) 

Now consider only

2x^3 - 9x^2 + 11x - 3

= x^2 * ( 2x - 9 ) + 1 * ( 11x - 3 )

The groups( 1.  1 * ( 11x - 3 ) and 2.  x^2 * ( 2x - 9 ) ) have no common factor and can not be added up to form a multiplication.

By polynomial long division process, we can write

2x^3 − 9x^2 + 11x − 3 = ( 2x −3 ) ( x^2 − 3x + 1 )

Now find the factors of x^2 − 3x + 1

According to the Quadratic Formula, x, the solution for  Ax^2+Bx+C  = 0  , where  A, B, and  C numbers, often called coefficients, is given by :

                                     x = (- B  ±  √ B^2 - 4AC) / 2A

Therefore, x = (- (-3) ± √( (-3)^2 - 4*1*1 ) ) / 2*1

x = ( -3 ± √( 9 - 4 ) ) / 2

x = ( -3 ± √5) / 2

Therefore the zeroes of f'(x) are ( 2x - 3 ), ( -3 + √5) / 2 and ( -3 - √5) / 2.

we can write 2x-3 = 0 i.e. x = 3/2 . so x = 1.5

From  ( -3 + √5) / 2 and ( -3 - √5) / 2, the x values are 2.61803 and  0.38197 respectively.

The real zeroes for f'(x) are :

x1 = 0.38197

x2 = 1.5

x3 = 2.61803

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