Consider the function f (x) = (2/3) x^3 + bx^2 + 2x + 3, where b is some undetermined coefficient: (a) find f'(x) and f''(x) and (b) if you know that f(x) has a stationary point at x = 2, use this information to find b.
(a) We have f(x)=(2/3)x3+ bx2 + 2x +3. First we have to solve for f'(x)=(dy/dx)f(x) which gives:f'(x) = [(32)/3]x2 + 2bx + 2 <=> f'(x)=2x2+ 2bx + 2 From there we do a second differentiation to solve for f''(x)=(dy/dx)f'(x) which gives:f''(x)= 4x + 2b Thus we have that f'(x)=2x2+ 2bx + 2 and f''(x)= 4x + 2b
(b) Since we know that f(x) has a stationary point at x=2, this implies that f'(x=2)=0, thus we have:f'(x=2)=222+2b*2+2 = 0 <=> f'(2) = 8 + 4b + 2 = 0 <=> f'(2) = 10 + 4b = 0, so rearranging to solve for b we get:10 + 4b = 0 <=> 10 = - 4b <=> b = - 10/4 which finally simplifies (if divided by 2) to:b = - 5/2
