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avanturin [10]
3 years ago
8

Is {(1,-2),(-2,0),(-1,2),(1,3)} a function

Mathematics
1 answer:
svp [43]3 years ago
5 0

Answer:

Step-by-step explanation:

No...this is not a function.

A function will not have ANY repeating x values.....they can have repeating y values, just not the x ones.

And this set of numbers has repeating x values of 1

in other words, in a function, all the x values have to be different numbers....none can be the same

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Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that
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The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)=f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........

Where the terms in f prime f^{1}(a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

f^{1}(a)=f^{1}(-4)=-8/(-4)^{2}

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The series of f(x) is as under:

f(x)=f(-4)+f^{1}(-4)(x+4)/1!+  f^{11}(-4)(x+4)^{2}/2!.............

=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+  24/(-4)^{3} (-4)(x+4)^{2}/2!.............

=-2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Learn more about taylor series at brainly.com/question/23334489

#SPJ4

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