Question

Find the local maximum and minimum values of f using both the First and Second Derivative Tests.
f(x) = 4 + 12x^2 − 8x^3

Answer 1

\[f(x)=4+12x ^{2}-8x ^{3}\]
\[f'(x)=24x-24x ^{2}=24x \left( 1-x \right)\]
f'(x)=0 gives x=0,1
f''(x)=24-48x
atx=0,f''(x)=24>0,there is local minima at x=0 and minimum value of f(x)=4
at x=1,f''(x)=-24<0,there is local maxima at x=1 and maximum value=4+12-8=8

Answer 2

first derivative test
when x<0 slightly
f'(x)=-(+)=-ve
when x>0 slightly
f'(x)=+(+)=+ve
f'(x) changes sign from -ve to +ve as x passes through 0,hence there is local minima at x=0
when x<1 slightly,x-1<0 or1-x>0
f'(x)=+(+)=+ve
when x>1 slightly,x-1>0 or 1-x<0
f'(x)=+(-)=-ve
f'(x) changes sign from +ve to-ve as x passes through 1,hence there is local maxima at x=1