Question

consider the function ( f(x)=1 - 3x^{2},-4leq xleq2 ).
the absolute maximum value is
and this occurs at ( x=)
the absolute minimum value is
and this occurs at ( x=)

Explanation:

Step1: Find the derivative of the function

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

Step2: Find the critical points

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

Step3: Evaluate the function at critical points and endpoints

• For \(x=-4\): \(f(-4)=1-3\times(-4)^{2}=1 - 48=-47\).
• For \(x = 0\): \(f(0)=1-3\times0^{2}=1\).
• For \(x = 2\): \(f(2)=1-3\times2^{2}=1-12=-11\).

Answer:

The absolute maximum value is \(1\) and this occurs at \(x = 0\).
The absolute minimum value is \(-47\) and this occurs at \(x=-4\).