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.单项选择题
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.单项选择题

类似问题

Find the absolute (global) maximum \( y \) value of the function:\( f(x)=2x^3+0x^2+-24x+5 \)On the interval \( [-9, 10] \)

Find the absolute (global) maximum [math: y] value of the function:[math: f(x)=2x3+0x2+−24x+5]On the interval [math: [−9,10]]

Maxima and Minima Graph the function Find the value of at which the given has a local maximum. Round to the nearest hundredth. [Fill in the blank] Find the value of at which the given has a local minimum. Round to the nearest hundredth. [Fill in the blank]

Find the extreme values of the function and where they occur. y =

What is the absolute maximum value of f ( x ) = x 2 x − 7 3 {"version":"1.1","math":"f(x)=x^2\sqrt[3]{x-7}"} on [ − 4 , 8 ] ? {"version":"1.1","math":"[-4,8]?"}

For the function f \colon \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = (4-x)e^x, consider its global maximum. This global maximum occurs at which value of x?

Find the absolute minimum and absolute maximum values of 𝑓 ( 𝑥 ) = 𝑥 3 − 3 𝑥 + 2 on the closed interval [ − 2 , 3 ] .

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单