To analyze the behavior of the function [tex]\( f(x) = -3x^2 - 18x - 26 \)[/tex], we start by finding its first and second derivatives.
1. First Derivative:
The first derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[
f'(x) = \frac{d}{dx} \left( -3x^2 - 18x - 26 \right) = -6x - 18
\][/tex]
2. Critical Points:
Critical points occur where [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[
-6x - 18 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
-6x = 18 \implies x = -3
\][/tex]
3. Intervals of Increase and Decrease:
Determine the sign of [tex]\( f'(x) \)[/tex] on the intervals around the critical point [tex]\( x = -3 \)[/tex]:
- For [tex]\( x < -3 \)[/tex] (e.g., [tex]\( x = -4 \)[/tex]):
[tex]\[
f'(-4) = -6(-4) - 18 = 24 - 18 = 6 > 0
\][/tex]
The function is increasing on [tex]\( (-\infty, -3) \)[/tex].
- For [tex]\( x > -3 \)[/tex] (e.g., [tex]\( x = -2 \)[/tex]):
[tex]\[
f'(-2) = -6(-2) - 18 = 12 - 18 = -6 < 0
\][/tex]
The function is decreasing on [tex]\( (-3, \infty) \)[/tex].
4. Second Derivative and Concavity:
The second derivative of the function [tex]\( f(x) \)[/tex] is:
[tex]\[
f''(x) = \frac{d}{dx} \left( -6x - 18 \right) = -6
\][/tex]
Since [tex]\( f''(x) < 0 \)[/tex] for all [tex]\( x \)[/tex], the function is concave down everywhere, indicating that any critical point will be a local maximum.
Evaluate [tex]\( f(x) \)[/tex] at the critical point [tex]\( x = -3 \)[/tex]:
[tex]\[
f(-3) = -3(-3)^2 - 18(-3) - 26 = -3(9) + 54 - 26 = -27 + 54 - 26 = 1
\][/tex]
Hence, there is a local maximum at [tex]\( (x, f(x)) = (-3, 1) \)[/tex].
Select the correct choices:
- The function is increasing on [tex]\( (-\infty, -3) \)[/tex]:
A. The function is increasing on [tex]\( (-\infty, -3) \)[/tex].
- The function is decreasing on [tex]\( (-3, \infty) \)[/tex]:
A. The function is decreasing on [tex]\( (-3, \infty) \)[/tex].
- The function has a local maximum at [tex]\( x = -3 \)[/tex] with [tex]\( f(x) = 1 \)[/tex]:
C. The function has a local maximum at [tex]\( x = -3 \)[/tex], [tex]\( f(x) = 1 \)[/tex], and no local minimum.
