Question

consider the function ( f(x)=1 - 7x^{2},-5leq xleq1 ).
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 \(f(x)=1 - 7x^{2}\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=(1)^\prime-(7x^{2})^\prime\). Since \((1)^\prime = 0\) and \((7x^{2})^\prime=14x\), we have \(f^\prime(x)=-14x\).

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 \([-5,1]\).

Step3: Evaluate the function at the critical point and endpoints

• Evaluate \(f(x)\) at \(x=-5\): \(f(-5)=1-7\times(-5)^{2}=1 - 175=-174\).
• Evaluate \(f(x)\) at \(x = 0\): \(f(0)=1-7\times0^{2}=1\).
• Evaluate \(f(x)\) at \(x = 1\): \(f(1)=1-7\times1^{2}=1 - 7=-6\).

Answer:

The absolute maximum value is \(1\) and this occurs at \(x = 0\). The absolute minimum value is \(-174\) and this occurs at \(x=-5\).