Question

find the global maximum and minimum points of the function $f(x)=3x^5 - 5x^3 + 27$ on the interval $-2,3$.
the global minimum is 25 at $x = 1$. the global maximum is 29 at $x = -1$.
the global minimum is 25 at $x = 1$. the global maximum is 27 at $x = 0$.
the global minimum is -29 at $x = -2$. the global maximum is 621 at $x = 3$.
the global minimum is 27 at $x = 0$. the global maximum is 29 at $x = -1$.

Explanation:

Step1: Find critical points via derivative

First, compute the derivative of $f(x)$:
$$f'(x) = 15x^4 - 15x^2 = 15x^2(x^2 - 1) = 15x^2(x-1)(x+1)$$
Set $f'(x)=0$, solve for $x$:
$15x^2(x-1)(x+1)=0$ gives $x=-1, 0, 1$. All are in $[-2,3]$.

Step2: Evaluate $f(x)$ at critical points

Calculate $f(x)$ at $x=-1,0,1$:

• $f(-1)=3(-1)^5 -5(-1)^3 +27 = -3 +5 +27 = 29$
• $f(0)=3(0)^5 -5(0)^3 +27 = 27$
• $f(1)=3(1)^5 -5(1)^3 +27 = 3 -5 +27 = 25$

Step3: Evaluate $f(x)$ at interval endpoints

Calculate $f(x)$ at $x=-2,3$:

• $f(-2)=3(-2)^5 -5(-2)^3 +27 = 3(-32) -5(-8) +27 = -96 +40 +27 = -29$
• $f(3)=3(3)^5 -5(3)^3 +27 = 3(243) -5(27) +27 = 729 -135 +27 = 621$

Step4: Compare all function values

List all values: $f(-2)=-29$, $f(-1)=29$, $f(0)=27$, $f(1)=25$, $f(3)=621$.
The smallest value is $-29$ (at $x=-2$), the largest is $621$ (at $x=3$).

Answer:

The global minimum is -29 at x = -2. The global maximum is 621 at x = 3.