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)=x² - 6x - 4; 0,4
find the first derivative of f.
f(x)=□
(simplify your answer.)
the absolute maximum value is □ at x = □.
(use a comma to separate answers as needed.)
the absolute minimum value is □ at x = □.
(use a comma to separate answers as needed.)

Explanation:

Step1: Find the first derivative

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), for \( f(x)=x^{2}-6x - 4 \), we have \( f^\prime(x)=(x^{2})^\prime-(6x)^\prime-(4)^\prime \).
Since \( (x^{2})^\prime = 2x \), \( (6x)^\prime=6 \), and \( (4)^\prime = 0 \), then \( f^\prime(x)=2x-6 \).

Step2: Find the critical points

Set \( f^\prime(x)=0 \), so \( 2x - 6=0 \).
Solving for \( x \):
Add \( 6 \) to both sides: \( 2x=6 \).
Divide both sides by \( 2 \): \( x = 3 \).

Step3: Evaluate the function at the critical point and endpoints

• For \( x = 0 \): \( f(0)=0^{2}-6\times0 - 4=-4 \).
• For \( x = 3 \): \( f(3)=3^{2}-6\times3 - 4=9 - 18 - 4=-13 \).
• For \( x = 4 \): \( f(4)=4^{2}-6\times4 - 4=16 - 24 - 4=-12 \).

Answer:

The first derivative \( f^\prime(x)=2x - 6 \).
The absolute maximum value is \(-4\) at \(x = 0\).
The absolute minimum value is \(-13\) at \(x = 3\).