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denpristay [2]
2 years ago
5

If p is a polynomial show that lim x→ap(x)=p(a

Mathematics
1 answer:
Lostsunrise [7]2 years ago
3 0

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

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AleksAgata [21]

Answer:

t≤4

Step-by-step explanation:

t+6≤10

Subtract 6 from each side

t+6-6≤10-6

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Let a graph have vertices {L, M, N, O, P, Q, R, S} and edge set {{L,R}, {M,P}, {M,Q}, {N,Q}, {P,R}, {Q,S}, {R,S}} .
Verdich [7]

Answer:

a) The degree of vertex P is 2.

b) The degree of vertex O is 0.

c) The graph has 2 components.

Step-by-step explanation:

a) The edges that have P as a vertice are {M,P} and {P,R}.

b) There is no edge with extreme point O.

c) One of the components is the one with the only vertex as O and has no edges. The other component is the one with the rest of the vertices and all the edges described.

The file has a realization of the graph.

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Use the properties of exponents to simplify the expression
ExtremeBDS [4]

Option B is correct after applying the property of integer exponent and simplification.

<h3>What is an integer exponent?</h3>

In mathematics, integer exponents are exponents that should be integers. It may be a positive or negative number. In this situation, the positive integer exponents determine the number of times the base number should be multiplied by itself.

We have:

=\rm (x.x^{-3}.y^{\dfrac{1}{3}})^2

=\rm (x^{-2}.y^{\dfrac{1}{3}})^2\\\\=\rm (x^{-4}.y^{\dfrac{2}{3}})\\\\=\rm \dfrac{y^{\dfrac{2}{3}}}{x^{4}}\\\\

Thus, option B is correct after applying the property of integer exponent and simplification.

Learn more about the integer exponent here:

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Plot all ordered pairs for the values in the domain D: (-8, -4, 0, 2, 6)
alexdok [17]

All ordered pairs for the values in the domain will be (-8, -3), (-4, -1), (0, 1), (2, 2), and (6, 4).

<h3>What is coordinate geometry?</h3>

Coordinate geometry is the study of geometry using the points in space.

The function is given below.

y = (1/2) x + 1

Then the domain is D: (-8, -4, 0, 2, 6)

Then the ordered pair will be

For x = -8

y = (1/2)(-8) + 1

y = -3

For x = -4

y = (1/2)(-4) + 1

y = -1

For x = 0

y = (1/2)(0) + 1

y = 1

For x = 2

y = (1/2)(2) + 1

y = 2

For x = 6

y = (1/2)(6) + 1

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All ordered pairs for the values in the domain will be (-8, -3), (-4, -1), (0, 1), (2, 2), and (6, 4).

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3 0
1 year ago
Will mark as brainlest PLz help
rosijanka [135]

Answer:

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