The least value of the function `f(x)=x^3-18 x^2+96 x` in the interval `[0,9]` is 126 (b) 135 (c) 160 (d) 0

Find the least value of the function f(x)=x^3-18 x^2+96 x in the interval [0,9] is ?

The least value of the function f(x)=x^(3)-18x^(2)+96x in the interval [0,9] is 126(b)135(c)160(d)0

The least value of the function f(x) = 2 cos x + x in the closed interval [0,pi/2] is

The Minimum value of the function f(x)=x^(3)-18x^(2)+96x in [0,9]

The least value of the function f(x) = 2 cos x + x in the closed interval [0, (pi)/(2)] is:

In Rolle's theorem the value of c for the function f(x)=x^(3)-3x in the interval [0,sqrt(3)] is

The value of c in Rolles theorem for the function f(x)=x^3-3x in the interval [0,\ sqrt(3)] is (a) 1 (b) -1 (c) 3//2 (d) 1//3

The maximum value of the function f(x)=((1+x)^(0. 6))/(1+x^(0. 6)) in the interval [0,1] is 2^(0. 4) (b) 2^(-0. 4) 1 (d) 2^(0. 6)

The maximum value of the function f(x)=((1+x)^(0. 6))/(1+x^(0. 6)) in the interval [0,1] is 2^(0. 4) (b) 2^(-0. 4) 1 (d) 2^(0. 6)

The least value of the function f(x) = ax + (b)/(x) (x gt 0, a gt 0, b gt 0)