Question

find the exact location of all the relative and absolute extrema of the function (f(x)=3x^{2}-6x - 9) with domain (0,3). (f) has ? at ((x,y)=) (smallest (x - value)) (f) has ? at ((x,y)=) (largest (x - value))

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=3x^{2}-6x - 9\). Using the power - rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=6x-6\).

Step2: Set the derivative equal to zero and solve for \(x\)

Set \(f^\prime(x)=0\), so \(6x - 6=0\). Add 6 to both sides: \(6x=6\), then divide by 6 to get \(x = 1\).

Step3: Evaluate the function at the critical point and endpoints

The endpoints of the domain \([0,3]\) are \(x = 0\) and \(x = 3\), and the critical point is \(x = 1\).

• When \(x = 0\), \(f(0)=3\times0^{2}-6\times0 - 9=-9\).
• When \(x = 1\), \(f(1)=3\times1^{2}-6\times1 - 9=3 - 6 - 9=-12\).
• When \(x = 3\), \(f(3)=3\times3^{2}-6\times3 - 9=27-18 - 9=0\).

Answer:

\(f\) has the smallest \(y - value\) at \((x,y)=(1,-12)\) (smallest \(x - value\) among the extrema is \(x = 1\)), \(f\) has the largest \(y - value\) at \((x,y)=(3,0)\) (largest \(x - value\) among the extrema is \(x = 3\))