Let
f(x)=3x4−8x3+12x2−48x+1 be a function defined on the interval
[1,4] First we have to find f'(x)
⇒f′(x)=12x3−24x2+24x−48 Now let's find the roots of the equation
f′(x)=0 ⇒12x3−24x2+24x−48=0 ⇒12.(x3−2x2+2x−4)=0 ⇒12.(x2+2).(x−2)=0 ⇒x−2=0[∵(x2+2)≠0. as we are dealing with real valued functions
] ⇒x=2 Now let's find out
fprime‌prime(x) i.e
⇒f"(x)=36x2−48x+24 Now evaluate the value of
f"(x) at
x=2, we get
⇒f"(2)=72>0 As we know that according to second derivative test if f''(c) > 0 then x = c is a point of local minima
So, x = 2 is a point of local minima
So f(2) = - 63
Let's calculate f(0) and f(3)
⇒ f(0) = 1 and f(3) = -8
As we know that if a function is defined in the closed interval [a, b] then the minimum value of f(x) on [a, b] is the smallest of m, f(a) and f(b).
So, the minimum value of the given function on the interval [0, 3] is - 63.