In the usual sense, it has no inverse function. Since it is not an injection (f(0)=f(2)=0), an inverse function would have to map 0 both to 0 and 2, and that’s impossible.

However, if you restrict the domain to [1,+oo> (to make it injective), and the codomain to [-1,+oo> (to make it surjective), then it has an inverse function, given by g(y)=1+sqrt(y+1).

6 0

3 years ago

Show that if S1 and S2 are subsets of a vector space V such that S1 c S2 then span(S1) c span(S2). In particular, if S1 c S2 the

klemol [59]

Answer:

See proof below

Step-by-step explanation:

Assume that V is a vector space over the field F (take F=R,C if you prefer).

Let $x\in span(S_1)$ . Then, we can write x as a linear combination of elements of s1, that is, there exist $v_1,v_2,\cdots,v_k \in S_1$ and $a_1,a_2,\cdots,a_k\in F$ such that $x=a_1v_1+a_2v_2+\cdots+a_kv_k$ . Now, $S_1\subseteq S_2$ then for all $y\in S_1$ we have that $y\in S_2$ . In particular, taking $y=v_j$ with $j=1,2,\cdots,k$ we have that $v_j\in S_2$ . Then, x is a linear combination of vectors in S2, therefore $x\in span(S_2)$ . We conclude that $span(S_1)\subseteq span(S_2)$ .

If, additionally $S_2\subseteq S_1$ then reversing the roles of S1 and S2 in the previous proof, $span(S_2)\subseteq span(S_1)$ . Then $span(S_1)\subseteq span(S_2)\subseteq span(S_1)$ , therefore $span(S_1)=span(S_2)$ .

5 0

3 years ago

What is the absolute value of |4.7|

larisa86 [58]

Answer:

4.7

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

113 to 1dpis it just 114?​

113 to 1dpis it just 114?