Question

consider the function $f(x)=3 - 3x^{2}$, $-4 < x < 1$.
the absolute maximum value is
and this occurs at $x$ equals
the absolute minimum value is
and this occurs at $x$ equals

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=3 - 3x^{2}\), and its derivative \(f^\prime(x)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\) is \(f^\prime(x)=-6x\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(-6x=0\), which gives \(x = 0\).

Step3: Evaluate the function at the critical point and endpoints

• For \(x=-4\): \(f(-4)=3-3\times(-4)^{2}=3 - 48=-45\).
• For \(x = 0\): \(f(0)=3-3\times0^{2}=3\).
• For \(x = 1\): \(f(1)=3-3\times1^{2}=0\).

Answer:

The absolute maximum value is \(3\) and this occurs at \(x = 0\). The absolute minimum value is \(-45\) and this occurs at \(x=-4\).