The derivative of f is given f'(x) = e^x(-x^3+3x)-3 for [0,5].

Question

What value of f(x) is an absolute minimum?

Answer 1

The  derivative of f is given f¹(x) = e^x(-x³ + 3x) - 3 for interval [0 , 5]

f1(x) = e¹(-x³ + 3x) - 3

Absolute minimum value  of f(x)  : put x = o in  f¹(x)

f¹(0) = e⁰(-0³ + 3(0)) - 3

f¹(0) = 1(0 + 0) - 3

f¹(0) = 0 - 3

f¹(0) = -3

There fore absolute minimum value of f(x) : -3.