Answer :
We want to estimate the local minimum for the function
[tex]$$
y = -x^6 - 6x^5 + 50x^3 + 45x^2 - 108x - 108.
$$[/tex]
A good strategy is to first find the critical points of the function by computing its derivative and setting it equal to zero. Then, by calculating the second derivative at these points, we can determine which of the critical points is a local minimum using the Second Derivative Test.
1. Find the Critical Points:
a. Compute the first derivative:
[tex]$$
y' = \frac{d}{dx}\left(-x^6 - 6x^5 + 50x^3 + 45x^2 - 108x - 108\right).
$$[/tex]
b. Setting [tex]$y' = 0$[/tex] produces an equation whose real solutions are the critical points. The numerical solving yields the following distinct real solutions:
[tex]$$
x = -3,\quad x \approx -1.618,\quad x \approx 0.618,\quad x = 2.
$$[/tex]
2. Determine the Nature of Each Critical Point:
a. Compute the second derivative [tex]$y''$[/tex]. The Second Derivative Test tells us:
• If [tex]$y''(x) > 0$[/tex], the function is concave up at that point and there is a local minimum.
• If [tex]$y''(x) < 0$[/tex], the function is concave down and there is a local maximum.
b. Evaluating [tex]$y''$[/tex] at the critical points shows:
• At [tex]$x \approx 0.618$[/tex], we have [tex]$y'' > 0$[/tex], indicating that this critical point is a local minimum.
• At the other critical points, either the second derivative is negative or the test is inconclusive (in the case of a duplicate or inflection point).
3. Evaluate the Function at the Local Minimum:
At [tex]$x \approx 0.618$[/tex], substituting back into the function gives
[tex]$$
y \approx -146.353.
$$[/tex]
Thus, the estimated local minimum is at the point
[tex]$$
\boxed{(0.618, -146.353)}.
$$[/tex]
Among the provided options, the correct answer is option D.
[tex]$$
y = -x^6 - 6x^5 + 50x^3 + 45x^2 - 108x - 108.
$$[/tex]
A good strategy is to first find the critical points of the function by computing its derivative and setting it equal to zero. Then, by calculating the second derivative at these points, we can determine which of the critical points is a local minimum using the Second Derivative Test.
1. Find the Critical Points:
a. Compute the first derivative:
[tex]$$
y' = \frac{d}{dx}\left(-x^6 - 6x^5 + 50x^3 + 45x^2 - 108x - 108\right).
$$[/tex]
b. Setting [tex]$y' = 0$[/tex] produces an equation whose real solutions are the critical points. The numerical solving yields the following distinct real solutions:
[tex]$$
x = -3,\quad x \approx -1.618,\quad x \approx 0.618,\quad x = 2.
$$[/tex]
2. Determine the Nature of Each Critical Point:
a. Compute the second derivative [tex]$y''$[/tex]. The Second Derivative Test tells us:
• If [tex]$y''(x) > 0$[/tex], the function is concave up at that point and there is a local minimum.
• If [tex]$y''(x) < 0$[/tex], the function is concave down and there is a local maximum.
b. Evaluating [tex]$y''$[/tex] at the critical points shows:
• At [tex]$x \approx 0.618$[/tex], we have [tex]$y'' > 0$[/tex], indicating that this critical point is a local minimum.
• At the other critical points, either the second derivative is negative or the test is inconclusive (in the case of a duplicate or inflection point).
3. Evaluate the Function at the Local Minimum:
At [tex]$x \approx 0.618$[/tex], substituting back into the function gives
[tex]$$
y \approx -146.353.
$$[/tex]
Thus, the estimated local minimum is at the point
[tex]$$
\boxed{(0.618, -146.353)}.
$$[/tex]
Among the provided options, the correct answer is option D.
