Question at position 8 On the interval [-1,+1], the function y = 4 + x2 -x3 has an absolute maximum when x =-0.50.51.-1.0.Single choice
A
-0.5
B
0.5
C
1.
D
-1.
E
0.

Question

Question at position 8 On the interval [-1,+1], the function y = 4 + x2 -x3 has an absolute maximum when x =-0.50.51.-1.0.

BlackTom AI · Accepted Answer

To determine where the absolute maximum occurs for y = 4 + x^2 - x^3 on the interval [-1, 1], we first identify critical points by differentiating the function.

1) Compute derivative: dy/dx = 2x - 3x^2 = x(2 - 3x).
Critical points come from setting dy/dx = 0, which gives x = 0 or x = 2/3. Both are within [-1, 1].

2) Evaluate the function at critical points and endpoints:
- At x = -1: y = 4 + (-1)^2 - (-1)^3 = 4 + 1 + 1 = 6.
- At x = 0: y = 4 + 0^2 - 0^3 = 4.
- At x = 2/3: y = 4 + (2/3)^2 - (2/3)^3 = 4 + 4/9 - 8/27 = 4 + 12/27 - 8/27 = 4 + 4/27 ≈ 4.148...
- At x = 1: y = 4 + 1^2 - 1^3 = 4 + 1 - 1 = 4.

3) Compare the values: 6 at x = -1 is the largest among {6, 4, ≈4.148, 4}.

Now assess each answer option:
-0.5: Plugging x = -0.5 gives y = 4 + 0.25 - (-0.125) = 4.25 + 0.125 = 4.375. This is not the maximum since it is less than 6, and higher values occur at x = -1.
0.5: For x = 0.5, y = 4 + 0.25 - 0.125 = 4.125. Again, not the maximum; it’s smaller than the value at -1 and even smaller than at 0 and 2/3.
1: At x = 1, y = 4 (as calculated). This is not the maximum either.
-1: At x = -1, y = 6, which is larger than all other evaluated points on the interval, making it the absolute maximum in this domain.
0: At x = 0, y = 4, which is not the maximum.

In summary, the values show that the absolute maximum on [-1, 1] occurs at x = -1 (value 6), while all other given candidate x-values yield smaller results. The incorrect options correspond to points that produce smaller y-values compared to the maximum found at x = -1.

Question at position 8 On the interval [-1,+1], the function y = 4 + x2 -x3 has an absolute maximum when x =-0.50.51.-1.0.Single choice

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