Taylor series question

can someone please explain to me how to get the Taylor Series for f(x)=1/6x-7

We want to get the denominator in the form a/(1 - r). To do this, divide the numerator and denominator by the constant in the denominator (-7) to get:
f(x) = 1/(6x - 7)
= (-1/7)/[(6x - 7)/(-7)]
= (-1/7)/(1 - 6x/7).

By the infinite geometric series:
sum(n=0 to infinity) x^n = 1/(1 - x).

Replacing x with 6x/7:
sum(n=0 to infinity) (6x/7)^n = 1/(1 - 6x/7)
==> sum(n=0 to infinity) (6^n * x^n)/7^n = 1/(1 - 6x/7).

Then, multiplying both sides by -1/7:
-1/7 * sum(n=0 to infinity) (6^n * x^n)/7^n = (-1/7)/(1 - 6x/7)
==> 1/(6x - 7) = sum(n=0 to infinity) -(6^n * x^n)/7^(n + 1).

This is the required Taylor Series.

I hope this helps!

This is wrong because your question is complete, you have not given a value of x.

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You need to give a value of x at which to base the Taylor series.