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Basile [38]
4 years ago
11

Use the given graph to determine the limit, if it exists.

Mathematics
2 answers:
Crank4 years ago
3 0

limit of f(x) as x approaches 3 from the right side.

x= 3.1 results in 3

x= 3.01 results in 3

x= 3.001 results in 3

limit as x approaches 3 from the right side is 3.

Kipish [7]4 years ago
3 0
<h2>Answer:</h2>

The  limit as x approaches three from the right of f of x is:

                                   3

<h2>Step-by-step explanation:</h2>

Based on the graph we observe that the graph of the function increases continuously from x= -∞ to x=3.

There is a open circle at (3,1)

Then at x=3 it takes the value as 7

i.e. when x=3 then f(3)=7

Now after x=3 i.e. to the right of x=3 the graph is a constant line i.e. f(x)=3 for x>3

There is a open circle at (3,3)

           Hence, the right hand limit of f(x) at x=3 is:

                     \lim_{x \to 3^+} f(x)=3

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<u>the answer to this inequality is:</u> x<6

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2 square root of 5 is between...<br> A) 2 and 3<br> B) 4 and 6<br> C) 6 and 9<br> D) 9 and 12
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7 0
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May somebody please help me with this question. It would mean a lot! Thank you
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6 0
4 years ago
If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?
Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

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