Question

find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x - values at which they occur
$f(x)=3x - 1,-4,5$
the absolute maximum value is \\( \square \\) at \\( x = \square \\)
(use a comma to separate answers as needed)
the absolute minimum value is \\( \square \\) at \\( x = \square \\)
(use a comma to separate answers as needed)

Explanation:

Step1: Find the derivative of the function

The function is \(f(x) = 3x - 1\). The derivative \(f^\prime(x)=\frac{d}{dx}(3x - 1)=3\). Since \(f^\prime(x)=3>0\) for all \(x\), the function is increasing on the interval \([-4,5]\).

Step2: Evaluate the function at the endpoints of the interval

• When \(x=-4\), \(f(-4)=3\times(-4)-1=-12 - 1=-13\).
• When \(x = 5\), \(f(5)=3\times5-1=15 - 1 = 14\).

Answer:

The absolute maximum value is \(14\) at \(x = 5\).
The absolute minimum value is \(-13\) at \(x=-4\).