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

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.