To determine the local maximum of the function [tex]\( y = x^4 - 2x^3 - 11x^2 + 12x + 36 \)[/tex], we need to follow these steps:
1. Find the first derivative of the function and set it to zero to find the critical points.
[tex]\[
f'(x) = \frac{d}{dx}(x^4 - 2x^3 - 11x^2 + 12x + 36)
= 4x^3 - 6x^2 - 22x + 12
\][/tex]
We need to solve [tex]\( 4x^3 - 6x^2 - 22x + 12 = 0 \)[/tex] for [tex]\( x \)[/tex].
2. Check the second derivative at the critical points to determine the nature (maxima or minima) of each critical point.
[tex]\[
f''(x) = \frac{d^2}{dx^2}(x^4 - 2x^3 - 11x^2 + 12x + 36)
= 12x^2 - 12x - 22
\][/tex]
3. Using the second derivative test: For each critical point [tex]\( x = c \)[/tex],
- If [tex]\( f''(c) < 0 \)[/tex], then [tex]\( f(c) \)[/tex] is a local maximum.
- If [tex]\( f''(c) > 0 \)[/tex], then [tex]\( f(c) \)[/tex] is a local minimum.
After solving these steps, you find the critical points of the function and evaluate them to identify any local maxima.
According to the solution derived, there is a critical point at [tex]\( x \approx 0.5 \)[/tex], where:
[tex]\[
f''(x) < 0 \quad \text{(Negative value, indicating a local maximum)}
\][/tex]
And,
[tex]\[
y = (0.5)^4 - 2(0.5)^3 - 11(0.5)^2 + 12(0.5) + 36
\approx 0.0625 - 0.25 - 2.75 + 6 + 36
\approx 39.0625
\][/tex]
Thus, the local maximum of the function is approximately at the point [tex]\( (0.5, 39.06) \)[/tex].
Therefore, the correct answer is:
C. [tex]\((0.5, 39.06)\)[/tex]
