How do you find the extrema of f(x)=4 x^3-26 x^2+16x+1f(x)=4x326x2+16x+1 on [0,3]?

1 Answer
May 19, 2015

The global maximum and minimum values of the continuous function ff on the closed and bounded (compact) interval [0,3][0,3] will occur either at the critical points inside the interval or at the endpoints.

The derivative is f'(x)=12x^2-52x+16=4(3x^2-13x+4)=4(3x-1)(x-4), which leads to critical points at x=1/3 and x=4. Only x=1/3 is in the interval [0,3], so that's the only one we need to consider.

Now f(0)=1, f(1/3)=4\cdot 1/27-26\cdot 1/9+16\cdot 1/3+1=97/27\approx 3.6, and f(3)=108-234+48+1=-77.

Therefore, the global maximum value of f on [0,3] is 97/27 at x=1/3 and the global minimum value of f on [0,3] is -77 at x=3.