# Given the function f(x)=(x-2)(x-3)(x-4)(x-5)+1, prove that f'(x)=0 has three real different roots.

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The function f(x) = (x - 2)(x - 3)(x - 4)(x - 5) + 1.

Use the product rule to find the derivative f'(x).

f'(x) = (x - 2)'(x - 3)(x - 4)(x - 5) + (x - 2)(x - 3)'(x - 4)(x - 5) + (x - 2)(x - 3)(x - 4)'(x - 5) + (x - 2)(x - 3)(x - 4)(x - 5)'

=> f'(x) = (x - 3)(x - 4)(x - 5) + (x - 2)(x - 4)(x - 5) + (x - 2)(x - 3)(x - 5) + (x - 2)(x - 3)(x - 4)

=> f'(x) = x^3 - 12x^2 + 47x - 60 + x^3 - 11x^2 + 38x - 40 + x^3 - 10x^2 + 31x - 30 + x^3 - 9x^2 + 26x - 24

=> f'(x) = 4x^3 - 42x^2 + 142x - 154

4x^3 - 42x^2 + 142x - 154 = 0

=> 2x^3 - 21x^2 + 71x - 77 = 0

=> 2x^3 - 7x^2 - 14x^2 + 49x + 22x - 77 = 0

=> x^2(2x - 7) - 7x(2x - 7) + 11( 2x - 7) = 0

=> (x^2 - 7x + 11)(2x - 7) = 0

x1 = 7/2

x2 = 7/2 + sqrt (49 - 44)/2

=> 7/2 + (sqrt 5)/2

x3 = 7/2 - (sqrt 5)/2

**The roots of f'(x) are 7/2 , 7/2 + (sqrt 5)/2 and 7/2 - (sqrt 5)/2. So we have 3 real and different roots.**

We'll differentiate the function with respect to x, applying the product rule:

f'(x) = (x-2)'*(x-3)*(x-4)*(x-5) + (x-2)*(x-3)'*(x-4)*(x-5) + (x-2)*(x-3)*(x-4)'*(x-5) + (x-2)*(x-3)*(x-4)*(x-5)'

f'(x) = (x-3)*(x-4)*(x-5) + (x-2)*(x-4)*(x-5) + (x-2)*(x-3)*(x-5) + (x-2)*(x-3)*(x-4)

f'(x) = (x-3)*(x-4)*(x - 2 + x - 5) + (x-2)*(x-5)*(x - 4 + x - 3)

f'(x) = (2x - 7)(x^2 - 7x + 10 +x^2 - 7x + 12)

We'll combine like terms inside brackets:

f'(x) = (2x - 7)(2x^2 - 14x + 22)

We'll cancel f'(x):

f'(x)=0

(2x - 7)(2x^2 - 14x + 22) = 0

We'll cancel each factor:

2x- 7 = 0

x = 7/2

2x^2 - 14x + 22 = 0 <=> x^2 - 7x + 11 = 0

We'll apply quadratic formula:

x1 = [7+sqrt(49 - 44)]/2

x1 = (7+sqrt5)/2

x2 = (7-sqrt5)/2

**The equation has 3 real roots and they are: {(7-sqrt5)/2; 7/2 ; (7+sqrt5)/2}**.

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