Answer :
The critical number within the interval [tex]\( (-\infty, A] \)[/tex] is A = -6.
To find the critical numbers, we need to find the points where the derivative of the function is equal to zero or undefined.
Given the function [tex]\( f(x) = 12x^5 + 45x^4 - 360x^3 + 5 \)[/tex], let's first find its derivative f'(x):
[tex]\[ f'(x) = 60x^4 + 180x^3 - 1080x^2 \][/tex]
Now, we'll set the derivative equal to zero and solve for x:
[tex]\[ 60x^4 + 180x^3 - 1080x^2 = 0 \][/tex]
We can factor out a common term:
[tex]\[ 60x^2(x^2 + 3x - 18) = 0 \][/tex]
Now, we'll solve [tex]\( x^2 + 3x - 18 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{{-3 \pm \sqrt{{3^2 - 4 \cdot 1 \cdot (-18)}}}}{{2 \cdot 1}} \][/tex]
[tex]\[ x = \frac{{-3 \pm 9}}{2} \][/tex]
So, we have two solutions:
[tex]\[ x_1 = \frac{{-3 + 9}}{2} = 3 \][/tex]
[tex]\[ x_2 = \frac{{-3 - 9}}{2} = -6 \][/tex]
Now, to determine which critical number falls within the interval [tex]\( (-\infty, A] \)[/tex], we need to evaluate the second derivative at these points.
The second derivative of f(x) is:
[tex]\[ f''(x) = 240x^3 + 540x^2 - 2160x \][/tex]
Let's evaluate f''(x) at x = -6 and x = 3:
For x = -6:
[tex]\[ f''(-6) = 240(-6)^3 + 540(-6)^2 - 2160(-6) = -12,960 \][/tex]
For \( x = 3 \):
[tex]\[ f''(3) = 240(3)^3 + 540(3)^2 - 2160(3) = 14,040 \][/tex]
Since the second derivative changes sign at x = -6, it indicates a local maximum. Therefore, the critical number within the interval [tex]\( (-\infty, A] \)[/tex] is A = -6.
