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Answer:
The minimum value of f(x) is 2
Step-by-step explanation:
- To find the minimum value of the function f(x), you should find the value of x which has the minimum value of y, so we will use the differentiation to find it
- Differentiate f(x) with respect to x and equate it by 0 to find x, then substitute the value of x in f(x) to find the minimum value of f(x)
∵ f(x) = 2x² - 4x + 4
→ Find f'(x)
∵ f'(x) = 2(2) - 4(1)
+ 0
∴ f'(x) = 4x - 4
→ Equate f'(x) by 0
∵ f'(x) = 0
∴ 4x - 4 = 0
→ Add 4 to both sides
∵ 4x - 4 + 4 = 0 + 4
∴ 4x = 4
→ Divide both sides by 4
∴ x = 1
→ The minimum value is f(1)
∵ f(1) = 2(1)² - 4(1) + 4
∴ f(1) = 2 - 4 + 4
∴ f(1) = 2
∴ The minimum value of f(x) is 2
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that . Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that . In this equality we can perform a right multiplication by
and obtain
. Then, in the obtained equality we perform a left multiplication by P and get
. If we write
and
we have
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and . So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.