4. Haitru Hanmenbulee H HaH60.nbluee 3Hauenns QYHKLHHH y=f(x) Ha orpeake a;b 26 y=x^5-5x^4+5x^3+1,[-1;2]

To find the maximum and minimum values of the function $y=f(x)$ on the interval $[-1;2]$, we need to find the critical points of the function within the interval and evaluate the function at the endpoints.<br /><br />The critical points occur where the derivative of the function is equal to zero or undefined. Let's find the derivative of the function $y=x^{5}-5x^{4}+5x^{3}+1$:<br /><br />$y' = 5x^{4} - 20x^{3} + 15x^{2}$<br /><br />Now, we set the derivative equal to zero and solve for $x$:<br /><br />$5x^{4} - 20x^{3} + 15x^{2} = 0$<br /><br />$5x^{2}(x^{2} - 4x + 3) = 0$<br /><br />$x^{2}(x - 1)(x - 3) = 0$<br /><br />The critical points are $x = 0$, $x = 1$, and $x = 3$. However, only $x = 0$ and $x = 1$ are within the interval $[-1;2]$.<br /><br />Now, we evaluate the function at the critical points and the endpoints:<br /><br />$f(-1) = (-1)^{5} - 5(-1)^{4} + 5(-1)^{3} + 1 = -1 - 5 - 5 + 1 = -10$<br /><br />$f(0) = (0)^{5} - 5(0)^{4} + 5(0)^{3} + 1 = 1$<br /><br />$f(1) = (1)^{5} - 5(1)^{4} + 5(1)^{3} + 1 = 1 - 5 + 5 + 1 = 2$<br /><br />$f(2) = (2)^{5} - 5(2)^{4} + 5(2)^{3} + 1 = 32 - 80 + 40 + 1 = -7$<br /><br />Therefore, the maximum value of the function $y=f(x)$ on the interval $[-1;2]$ is $2$, and the minimum value is $-10$.