Question

f(x) = (x-1)(x-2)(x-3) , x ε[0,4], find if 'c' LMVT can be applied

Answer 1

f(x) = (x-1)(x-2)(x-3) ,           x ∈[0,4],

∴ f(x) = x³ -6x² + 11x -6

As f(x) is a polynomial in x
(1) f(x) is continuous on [0, 4]
(2) f(x) is differentiable on (0, 4) 
Thus, all the conditions of LMVT are satisfied. 

To verify LMVT we have to find c ∈ (0,4) such that 

`"f'" ("c") = ("f"(4 )-"f"(0))/(4-0)`         ........(1)

Now `"f"(4) = (4-1)(4-2)(4-3) = 6`

f(0) = (0-1)(0-2)(0-3) = -6 and

f '(x) = 3x² - 12x +11

∴ f '(c) = 3c² -12c +11

∴ from (1)  

`("f"(4) -"f"(0))/(4-0) = 3"c"^2 - 12"c" +11`

`(6-(-6))/4 = 3"c"^2 -12 "c" +11`

3 = 3c² -12c +11

3c² -12c +8 =0

c = `("b"+-sqrt("b"^2 - 4"ac"))/"2a"`

c = `(12±sqrt(144 -96))/6`

c =`(12± sqrt48)/6`

c = `12 ±2/3 sqrt3`  

c =`  4 ± 2/sqrt3`

Both the value of c lie between 0 and 4.