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2 answers:
Answer:
x ∈ [-1, 2]
Step-by-step explanation:
The solution to the equation |x+1|+|x-2|=3 can be found by separately considering the domains where the function is defined differently.
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<h3>analysis</h3>The absolute value function y=|x| has a vertex (turning point) at x=0. The translated function y=|x-h| will have its turning point at x=h. The sum of two absolute value functions with different turning points will have two turning points: one at each of the turning points of the constituents.
|x+1| has a turning point at x=-1
|x-2| has a turning point at x=2
This means there are three different domains to consider when determining solutions to the equation: (-∞, -1), [-1, 2), [2, ∞).
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<h3>(-∞, -1)</h3>In this domain, the equation is equivalent to ...
-(x+1) -(x -2) = 3
-2x = 2 . . . . . . . . . simplify, add -1
x = -1 . . . . . . . . . . divide by -2, this x-value is not in the domain
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<h3>[-1, 2)</h3>In this domain, the equation is equivalent to ...
(x +1) -(x -2) = 3
3 = 3 . . . . . true for all x in the domain
The solution is x ∈ [-1, 2).
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<h3>[2, ∞)</h3>In this domain, the equation is equivalent to ...
(x +1) +(x -2) = 3
2x = 4 . . . . . . . . . . simplify, add 1
x = 2 . . . . . . . . . . . divide by 2. This x-value is in the domain
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<h3>consolidate solutions</h3>Our solutions are ...
x ∈ [-1, 2) ⋃ [2, 2]
or ...
x ∈ [-1, 2]
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<em>Additional comment</em>
The graph shows the solution to the equation f(x) = 0, where the given equation is used to define the function f(x):
f(x) = |x +1| +|x -2| -3
f(x) will be zero for values of x that satisfy the original equation. We note that the solution is the closed interval [-1, 2], as we found above.
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Answer:
Only d) is false.
Step-by-step explanation:
Let be the characteristic polynomial of B.
a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)
b) Remember that . 0 is a root of p, so we have that
.
c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.
d) det(B)=0 by part c) so B is not invertible.
e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.