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

The global maximum and minimum values of the continuous function $f$ on the closed and bounded (compact) interval $\left[0 , 3\right]$ will occur either at the critical points inside the interval or at the endpoints.
The derivative is $f ' \left(x\right) = 12 {x}^{2} - 52 x + 16 = 4 \left(3 {x}^{2} - 13 x + 4\right) = 4 \left(3 x - 1\right) \left(x - 4\right)$, which leads to critical points at $x = \frac{1}{3}$ and $x = 4$. Only $x = \frac{1}{3}$ is in the interval $\left[0 , 3\right]$, so that's the only one we need to consider.
Now $f \left(0\right) = 1$, $f \left(\frac{1}{3}\right) = 4 \setminus \cdot \frac{1}{27} - 26 \setminus \cdot \frac{1}{9} + 16 \setminus \cdot \frac{1}{3} + 1 = \frac{97}{27} \setminus \approx 3.6$, and $f \left(3\right) = 108 - 234 + 48 + 1 = - 77$.
Therefore, the global maximum value of $f$ on $\left[0 , 3\right]$ is $\frac{97}{27}$ at $x = \frac{1}{3}$ and the global minimum value of $f$ on $\left[0 , 3\right]$ is $- 77$ at $x = 3$.