Question

select the true statement about the extreme value of the function defined by the expression below 3x² - 12x - 36 a. the extreme value is a maximum at the point (-2,48). b. the extreme value is a maximum at the point (2,48). c. the extreme value is a minimum at the point (-2,-48). d. the extreme value is a minimum at the point (2,-48).

Explanation:

Step1: Find the derivative of the function

Let \(y = 3x^{2}-12x - 36\). The derivative \(y^\prime=\frac{d}{dx}(3x^{2}-12x - 36)=6x - 12\).

Step2: Find the critical points

Set \(y^\prime = 0\), so \(6x-12 = 0\). Solving for \(x\):
\(6x=12\), then \(x = 2\).

Step3: Find the second - derivative

The second - derivative \(y^{\prime\prime}=\frac{d}{dx}(6x - 12)=6\).

Step4: Use the second - derivative test

Since \(y^{\prime\prime}(2)=6>0\), when \(x = 2\), the function has a minimum.
When \(x = 2\), \(y=3\times(2)^{2}-12\times2 - 36=12-24 - 36=-48\).

Answer:

D. The extreme value is a minimum at the point \((2,-48)\)