Question

closed interval method: problem 2
(1 point)
consider the function ( f(x)=4 - 7x^{2}, - 3leq xleq1 ).
the absolute maximum value is
and this occurs at ( x ) equals
the absolute minimum value is
and this occurs at ( x ) equals
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Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=4 - 7x^{2}\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=-14x\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(-14x=0\), which gives \(x = 0\). The critical point \(x = 0\) lies within the interval \([-3,1]\).

Step3: Evaluate the function at the critical point and endpoints

• For \(x=-3\): \(f(-3)=4-7\times(-3)^{2}=4 - 63=-59\).
• For \(x = 0\): \(f(0)=4-7\times0^{2}=4\).
• For \(x = 1\): \(f(1)=4-7\times1^{2}=4 - 7=-3\).

Answer:

The absolute maximum value is \(4\) and this occurs at \(x\) equals \(0\). The absolute minimum value is \(-59\) and this occurs at \(x\) equals \(-3\).