To find the maximum or minimum value of the function [tex]\(f(s) = s^2 - 1.6s + 11\)[/tex], we need to follow a systematic approach using calculus. Here is a detailed, step-by-step solution:
1. Find the first derivative:
The first derivative of the function [tex]\(f(s)\)[/tex] helps us determine the critical points where the function might have a maximum or minimum.
[tex]\[
f'(s) = \frac{d}{ds}(s^2 - 1.6s + 11) = 2s - 1.6
\][/tex]
2. Set the first derivative equal to zero and solve for [tex]\(s\)[/tex]:
To find the critical points, solve the equation [tex]\(f'(s) = 0\)[/tex]:
[tex]\[
2s - 1.6 = 0
\][/tex]
[tex]\[
2s = 1.6
\][/tex]
[tex]\[
s = \frac{1.6}{2} = 0.8
\][/tex]
3. Find the second derivative:
The second derivative of the function will help us determine the concavity at the critical point, which tells us whether it is a maximum or minimum.
[tex]\[
f''(s) = \frac{d}{ds}(2s - 1.6) = 2
\][/tex]
4. Evaluate the second derivative at the critical point:
Substitute [tex]\(s = 0.8\)[/tex] into the second derivative:
[tex]\[
f''(0.8) = 2
\][/tex]
Since [tex]\(f''(0.8) > 0\)[/tex], the function is concave up at [tex]\(s = 0.8\)[/tex]. Thus, [tex]\(s = 0.8\)[/tex] is a point of local minimum.
5. Find the value of the function at the critical point:
Substitute [tex]\(s = 0.8\)[/tex] back into the original function to find the minimum value:
[tex]\[
f(0.8) = (0.8)^2 - 1.6(0.8) + 11
\][/tex]
[tex]\[
f(0.8) = 0.64 - 1.28 + 11
\][/tex]
[tex]\[
f(0.8) = 10.36
\][/tex]
Therefore, the minimum value of the function [tex]\( f(s) = s^2 - 1.6s + 11 \)[/tex] occurs at [tex]\( s = 0.8 \)[/tex] and the minimum value is [tex]\( f(0.8) = 10.36 \)[/tex].
To summarize:
- Critical point: [tex]\( s = 0.8 \)[/tex]
- Value at critical point: [tex]\( f(0.8) = 10.36 \)[/tex] (rounded to two decimal places)
- Nature of value: minimum value
