Question

1. (2 points) find the global maximum and minimum for the function, if they exist.

\\( f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + 5 \\) on \\( - 1 \leq x \leq 1 \\)
fully justify your answers and write the name of the theorem/test that you use.

Explanation:

Step1: Find the derivative

By the power rule, if \(y = x^n\), then \(y^\prime=nx^{n - 1}\). For \(f(x)=x^{3}-3x^{2}+5\), \(f^\prime(x)=3x^{2}-6x=3x(x - 2)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(3x(x - 2)=0\). Solving for \(x\), we get \(x = 0\) or \(x=2\). But \(x = 2
otin[-1,1]\), so the only critical point in the interval \([-1,1]\) is \(x = 0\) (by the definition of critical points: points where \(f^\prime(x)=0\) or \(f^\prime(x)\) is undefined. Here \(f^\prime(x)\) is a polynomial, so it is defined everywhere in the interval).

Step3: Evaluate the function at the critical point and endpoints

• Evaluate \(f(x)\) at \(x=-1\): \(f(-1)=(-1)^{3}-3(-1)^{2}+5=-1 - 3+5=1\)
• Evaluate \(f(x)\) at \(x = 0\): \(f(0)=0^{3}-3\times0^{2}+5=5\)
• Evaluate \(f(x)\) at \(x = 1\): \(f(1)=1^{3}-3\times1^{2}+5=1-3 + 5=3\)

We use the Extreme - Value Theorem (a continuous function \(y = f(x)\) on a closed interval \([a,b]\) attains its global maximum and minimum on \([a,b]\). Since \(f(x)=x^{3}-3x^{2}+5\) is a polynomial, it is continuous on \([-1,1]\))

Answer:

The global minimum is \(1\) (at \(x=-1\)) and the global maximum is \(5\) (at \(x = 0\))