+118653 whats the max of this function f(x)=(x^3/3)-4x^2+12x+1
The maximum is where the first derivative is 0 and the second derivative is less than 0.
\(f(x)=(x^3/3)-4x^2+12x+1\\ f'(x)=x^2-8x+12\\ f''(x)=2x-8\\~\\ x^2-8x+12=0\\ (x-2)(x-6)=0\\ x=+2\;\; or \;\;x=+6\\ y''(2)=4-8<0 \quad \text{Max at x=2}\\ y''(6)=12-8>0 \quad \text{Min at x=6}\\ When\;\;x=2\\ f(2)=8/3-4*4+12*2+1\\ f(2)=8/3-16+24+1\\ f(2)=8/3+9\\ f(2)=11\frac{2}{3}\\ \text{So there is one (local) maximum point at }(2,11\frac{2}{3}) \)
Of course the limit of this function as x tends to infinity is infinity so there are an infinite number of function values that are higher than this maximum. :)
