问题1.在不使用R上函数f(x)= 4x 2 – 4x + 4的导数的情况下，找到最大值和最小值(如果有)
解决方案：
Given in the question f(x) = 4x2 – 4x + 4 on R
= 4x2 – 4x + 1 + 3
If we group the above equation we will get,
= (2x – 1)2 + 3
⇒ (2x – 1)2 ≥ 0
= (2x – 1)2 + 3 ≥ 3
= f(x) ≥ f (½)
So we can say that, the minimum value of f(x) is 3 at x = ½
Since, f(x) can be infinite. So maximum value does not exist.
问题2。找到最大值和最小值，如果有的话，不使用R上f(x)= –(x – 1) 2 + 2给定的函数的导数
解决方案：
Given function in the question is f(x) = – (x – 1)2 + 2
We can observe that (x – 1)2 ≥ 0 ∀ x ∈ R
Therefore, f(x) = – (x – 1)2 + 2 ≤ 2 ∀ x ∈ R
The maximum value of function can be attained when
⇒ (x – 1) = 0
⇒ (x – 1) = 0, x = 1
Maximum value of function = f(1) = – (1 – 1)2 + 2 = 2
So we can say that function f does not have minimum value.
问题3.找出最大值和最小值(如果有的话)，而无需使用f(x)= | x + 2 |给出的函数的导数。在R上
解决方案：
Given function in the question is f(x) = |x + 2| on R
⇒ f(x) ≥ 0 ∀ x ∈ R
The minimum value of f(x) = 0, which can be attained at x = -2
So We can say that, f(x) = |x + 2| does not have the maximum value.
问题4.找出最大值和最小值(如果有)，而不使用R上f(x)= sin2x + 5给出的函数的导数
解决方案：
Given function in the question is f(x) = sin2x + 5 on R
As we know that range of sin2x is [-1,1].
Adding 5 both side.
⇒ – 1 + 5 ≤ sin2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
So we can say that, the maximum value is 6 and minimum value is 4.
问题5.找出最大值和最小值(如果有)，而无需使用f(x)= | sin 4x + 3 |给出的函数的导数在R上
解决方案：
Given function in the question is f(x) = |sin 4x + 3| on R
As we know that range of sin 4x is [-1,1]
Adding 3 both side:
⇒ 2 ≤ sin 4x + 3 ≤ 4
⇒ 2 ≤ |sin 4x + 3| ≤ 4
So we can say that, the maximum value is 4 and minimum value is 2.
问题6.找出最大值和最小值，如果不使用R上f(x)= 2x 3 + 5给出的函数的导数的话。
解决方案：
f(x) = 2x3 + 5 on R.
We can observe that the values of f(x) increased when the values of x are increased.
f(x) can be as large as possible.
So we can conclude that , f(x) does not have the maximum value.
Also, f(x) can be made small by giving smaller values to x.
So we can say that f(x) does not have the minimum value.
问题7.在不使用g(x)=-|给定的函数导数的情况下，找到最大值和最小值。 x + 1 | + 3
解决方案：
We have given this equation g(x) = -| x + 1 | + 3
As we know that -| x + 1 | ≤ 0 ∀ x ∈ R
⇒ g(x) = -| x + 1 | + 3 ≤ 3 ∀ x ∈ R
The maximum value of function g is attained at | x + 1 | = 0
| x + 1 | = 0
⇒ x = -1
⇒ Maximum value of g = g(-1) = – | -1 + 1 | + 3 = 3
So we can conclude that function g does not have minimum value.
问题8.找出最大值和最小值，如果不使用R上f(x)= 16x 2 – 16x + 28给定的函数的导数的话。
解决方案：
We have given this equation f(x) = 16x2 – 16x + 28 on R.
= 16x2 – 16x + 4 + 24
= (4x – 2)2 + 24
⇒ (4x – 2)2 ≥ 0 ∀ x ∈ R
⇒ (4x – 2)2 + 24 ≥ 24 ∀ x ∈ R
⇒ f(x) ≥ f(½)
So the minimum value of f(x) is 24 at x = ½
So we can say that, f(x) can be made as large as possible by giving different values to x.
So we can conclude that, maximum values does not exist.
问题9.找出最大值和最小值，如果不使用R上f(x)= x 3 – 1给定的函数的导数的话。
解决方案：
We have given this equation f(x) = f(x) = x3 – 1 on R.
We can observe that the values of f(x) increases when the values of x are increased.
f(x) can be made as large as possible by putting values of x.
So, we can say that f(x) does not have the maximum value.
Also, f(x) can be made as small as giving by smaller values to x.
So we can conclude that f(x) does not have the minimum value.