Question

if f(x) = 3x^3 - 3x^2 + 5, then find the interval where f(x) is concave up

Answer 1

\[f \prime \left( x \right)=9x^2-6x=3x \left( 3x-2 \right)\]
f'(x)=0 gives x=0,x=2/3
\[f''(x)=18x-6,when~x=0.f''(x)=-6<0,there~ is~ maxima~at~x=0\]
when x=2/3,
\[f''(x)=18*\frac{ 2 }{ 3}-6=12-6=6>0\]
There is minima at x=2/3

Answer 2

Refer to the Mathematica attachment.