Consider the function $f(x)=(10/x^2)-(7/x^6)$ if f(1)=0, then what is f(x)?

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Answer 1 · 2015-08-21T20:48:29

If I'm correct that that should be $f'(x)$ to start, then $f(x) = -10/x + 7/(5x^5) +43/5$

I'm going to guess the question is intended to be:

$f'(x)=(10/x^2)-(7/x^6)$ if $f(1)=0$ , then what is $f(x)$ ?

$f'(x)=10x^-2 - 7x^-6$

$f(x) = 10x^(-2+1)/(-2+1) - 7 x^(-6+1)/(-6+1) +C$

$= 10x^-1/-1-7x^-5/-5+C$

So,

$f(x) = -10/x + 7/(5x^5) +C$

Now, $f(1) = -10/1+7/(5(1))+C = 0$ gets us:

$C = 43/5$ and

$f(x) = -10/x + 7/(5x^5) +43/5$

Consider the function f(x)=(10/x^2)-(7/x^6)f(x)=(10/x^2)-(7/x^6) if f(1)=0, then what is f(x)?