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

Answer 1

Aurora Alvarado

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#

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Answer 2

Christopher Allers

If ( f(1) = 0 ), then ( f(x) = \frac{{10}}{{x^2}} - \frac{{7}}{{x^6}} ) becomes ( f(x) = \frac{{10}}{{x^2}} - \frac{{7}}{{x^6}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.