For the function 𝑓(𝑥)=(16/𝑥^2), simplify each expression as much as possible.












-2












$begingroup$


For the function 𝑓(𝑥)=(16/𝑥^2), simplify each expression as much as possible.




  1. (𝑓(𝑥+ℎ)−𝑓(𝑥))/ℎ, ℎ≠0:

  2. (𝑓(𝑤)−𝑓(𝑥))/𝑤−𝑥, 𝑥≠𝑤:










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    Welcome to MathSE. When you pose a question here, it is expected that you share your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
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    – N. F. Taussig
    Jan 19 at 22:59










  • $begingroup$
    What don't you understand? just write all the $f(stuff)$ ans $frac {16}{stuff^2}$ and then do all the math to simplify. It's ... work. But it's straight forwards and unless you tell us where you have problems there's no reason you shouldn't be able to do it.
    $endgroup$
    – fleablood
    Jan 20 at 0:37
















-2












$begingroup$


For the function 𝑓(𝑥)=(16/𝑥^2), simplify each expression as much as possible.




  1. (𝑓(𝑥+ℎ)−𝑓(𝑥))/ℎ, ℎ≠0:

  2. (𝑓(𝑤)−𝑓(𝑥))/𝑤−𝑥, 𝑥≠𝑤:










share|cite|improve this question









$endgroup$












  • $begingroup$
    Welcome to MathSE. When you pose a question here, it is expected that you share your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 19 at 22:59










  • $begingroup$
    What don't you understand? just write all the $f(stuff)$ ans $frac {16}{stuff^2}$ and then do all the math to simplify. It's ... work. But it's straight forwards and unless you tell us where you have problems there's no reason you shouldn't be able to do it.
    $endgroup$
    – fleablood
    Jan 20 at 0:37














-2












-2








-2





$begingroup$


For the function 𝑓(𝑥)=(16/𝑥^2), simplify each expression as much as possible.




  1. (𝑓(𝑥+ℎ)−𝑓(𝑥))/ℎ, ℎ≠0:

  2. (𝑓(𝑤)−𝑓(𝑥))/𝑤−𝑥, 𝑥≠𝑤:










share|cite|improve this question









$endgroup$




For the function 𝑓(𝑥)=(16/𝑥^2), simplify each expression as much as possible.




  1. (𝑓(𝑥+ℎ)−𝑓(𝑥))/ℎ, ℎ≠0:

  2. (𝑓(𝑤)−𝑓(𝑥))/𝑤−𝑥, 𝑥≠𝑤:







algebra-precalculus






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asked Jan 19 at 22:35









mfmmfm

11




11












  • $begingroup$
    Welcome to MathSE. When you pose a question here, it is expected that you share your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 19 at 22:59










  • $begingroup$
    What don't you understand? just write all the $f(stuff)$ ans $frac {16}{stuff^2}$ and then do all the math to simplify. It's ... work. But it's straight forwards and unless you tell us where you have problems there's no reason you shouldn't be able to do it.
    $endgroup$
    – fleablood
    Jan 20 at 0:37


















  • $begingroup$
    Welcome to MathSE. When you pose a question here, it is expected that you share your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 19 at 22:59










  • $begingroup$
    What don't you understand? just write all the $f(stuff)$ ans $frac {16}{stuff^2}$ and then do all the math to simplify. It's ... work. But it's straight forwards and unless you tell us where you have problems there's no reason you shouldn't be able to do it.
    $endgroup$
    – fleablood
    Jan 20 at 0:37
















$begingroup$
Welcome to MathSE. When you pose a question here, it is expected that you share your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 22:59




$begingroup$
Welcome to MathSE. When you pose a question here, it is expected that you share your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 22:59












$begingroup$
What don't you understand? just write all the $f(stuff)$ ans $frac {16}{stuff^2}$ and then do all the math to simplify. It's ... work. But it's straight forwards and unless you tell us where you have problems there's no reason you shouldn't be able to do it.
$endgroup$
– fleablood
Jan 20 at 0:37




$begingroup$
What don't you understand? just write all the $f(stuff)$ ans $frac {16}{stuff^2}$ and then do all the math to simplify. It's ... work. But it's straight forwards and unless you tell us where you have problems there's no reason you shouldn't be able to do it.
$endgroup$
– fleablood
Jan 20 at 0:37










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$begingroup$

$$frac{f(x+h)-f(x)}{h}=frac{frac{16}{(x+h)^2}-frac{16}{x^2}}{h}=frac{frac{16[x^2-(x+h)^2]}{x^2(x+h)^2}}{h}=frac{16(x-x-h)(x+x+h)}{hx^2(x+h)^2}=frac{16h(2x+h)}{hx^2(x+h)^2}=frac{16(2x+h)}{x^2(x+h)^2}=frac{16}{x(x+h)^2}+frac{16}{x^2(x+h)}$$



$$frac{f(w)-f(x)}{w-x}=frac{frac{16}{w^2}-frac{16}{x^2}}{w-x}=frac{16(x^2-w^2)}{w^2x^2(w-x)}=-frac{16(w+x)}{w^2x^2}$$






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    $begingroup$

    $$frac{f(x+h)-f(x)}{h}=frac{frac{16}{(x+h)^2}-frac{16}{x^2}}{h}=frac{frac{16[x^2-(x+h)^2]}{x^2(x+h)^2}}{h}=frac{16(x-x-h)(x+x+h)}{hx^2(x+h)^2}=frac{16h(2x+h)}{hx^2(x+h)^2}=frac{16(2x+h)}{x^2(x+h)^2}=frac{16}{x(x+h)^2}+frac{16}{x^2(x+h)}$$



    $$frac{f(w)-f(x)}{w-x}=frac{frac{16}{w^2}-frac{16}{x^2}}{w-x}=frac{16(x^2-w^2)}{w^2x^2(w-x)}=-frac{16(w+x)}{w^2x^2}$$






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      0












      $begingroup$

      $$frac{f(x+h)-f(x)}{h}=frac{frac{16}{(x+h)^2}-frac{16}{x^2}}{h}=frac{frac{16[x^2-(x+h)^2]}{x^2(x+h)^2}}{h}=frac{16(x-x-h)(x+x+h)}{hx^2(x+h)^2}=frac{16h(2x+h)}{hx^2(x+h)^2}=frac{16(2x+h)}{x^2(x+h)^2}=frac{16}{x(x+h)^2}+frac{16}{x^2(x+h)}$$



      $$frac{f(w)-f(x)}{w-x}=frac{frac{16}{w^2}-frac{16}{x^2}}{w-x}=frac{16(x^2-w^2)}{w^2x^2(w-x)}=-frac{16(w+x)}{w^2x^2}$$






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        $begingroup$

        $$frac{f(x+h)-f(x)}{h}=frac{frac{16}{(x+h)^2}-frac{16}{x^2}}{h}=frac{frac{16[x^2-(x+h)^2]}{x^2(x+h)^2}}{h}=frac{16(x-x-h)(x+x+h)}{hx^2(x+h)^2}=frac{16h(2x+h)}{hx^2(x+h)^2}=frac{16(2x+h)}{x^2(x+h)^2}=frac{16}{x(x+h)^2}+frac{16}{x^2(x+h)}$$



        $$frac{f(w)-f(x)}{w-x}=frac{frac{16}{w^2}-frac{16}{x^2}}{w-x}=frac{16(x^2-w^2)}{w^2x^2(w-x)}=-frac{16(w+x)}{w^2x^2}$$






        share|cite|improve this answer









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        $$frac{f(x+h)-f(x)}{h}=frac{frac{16}{(x+h)^2}-frac{16}{x^2}}{h}=frac{frac{16[x^2-(x+h)^2]}{x^2(x+h)^2}}{h}=frac{16(x-x-h)(x+x+h)}{hx^2(x+h)^2}=frac{16h(2x+h)}{hx^2(x+h)^2}=frac{16(2x+h)}{x^2(x+h)^2}=frac{16}{x(x+h)^2}+frac{16}{x^2(x+h)}$$



        $$frac{f(w)-f(x)}{w-x}=frac{frac{16}{w^2}-frac{16}{x^2}}{w-x}=frac{16(x^2-w^2)}{w^2x^2(w-x)}=-frac{16(w+x)}{w^2x^2}$$







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        answered Jan 19 at 22:46









        user289143user289143

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