Find the absolute minimum of the function $f ( x ) = x ^ { 3 } - 3 x ^ { 2 }$ on the interval -1$\le$ x $\le$ 3.
A) -1
B) -4
C) 0
D) 3

Question

Quizplus · Accepted Answer

To find the absolute minimum of the function $f(x) = x^3 - 3x^2$ on the interval $[-1, 3]$, we first find its critical points by taking the derivative and setting it equal to zero. The derivative is $f'(x) = 3x^2 - 6x$. Setting $f'(x) = 0$ gives $x^2 - 2x = 0$, which factors to $x(x - 2) = 0$, so the critical points are $x = 0$ and $x = 2$. Evaluating the function at these points and at the endpoints of the interval, we get:- $f(-1) = (-1)^3 - 3(-1)^2 = -1 - 3 = -4$- $f(0) = 0^3 - 3(0)^2 = 0$- $f(2) = 2^3 - 3(2)^2 = 8 - 12 = -4$- $f(3) = 3^3 - 3(3)^2 = 27 - 27 = 0$The minimum value of the function on the interval is $-4$, which occurs at $x = -1$ and $x = 2$.

Find the Absolute Minimum of the Function f(x)=x3−3x2f ( x ) = x ^ { 3 } - 3 x ^ { 2 }f(x)=x3−3x2

Find the Absolute Minimum of the Function $f ( x ) = x ^ { 3 } - 3 x ^ { 2 }$