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Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.