inverse function

[spacewater]

problem
\(\displaystyle f(x) = x^{-1}\)

\(\displaystyle \frac{f(x+h) - f(x)}{h}\)

steps
\(\displaystyle \frac{f(\frac {1}{x}+h)-f(\frac{1}{x})}{h}\)

\(\displaystyle \frac {h}{h} = 1\)

final answer
\(\displaystyle 1\)

 

[Deleted member 4993]

problem
\(\displaystyle f(x) = x^{-1}\)

\(\displaystyle \frac{f(x+h) - f(x)}{h}\)

steps
\(\displaystyle \frac{f(\frac {1}{x}+h)-f(\frac{1}{x})}{h}\)

\(\displaystyle \frac {h}{h} = 1\)

final answer
\(\displaystyle 1\)

I don't know - what you are trying to do - it is so confused it is not even wrong!

\(\displaystyle f(x) \, = \, \frac{1}{x}\)

\(\displaystyle f(x+h) \, = \, \frac{1}{x+h}\)

\(\displaystyle \frac{f(x+h)-f(x)}{h} \, = \, \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \, = \, \frac{x - (x+h)}{h\cdot x \cdot (x+h)}\)

Now finish it ... carefully

 

[spacewater]

problem
\(\displaystyle f(x) = x^{-1}\)

\(\displaystyle \frac{f(x+h) - f(x)}{h}\)

steps
\(\displaystyle \frac{f(\frac {1}{x}+h)-f(\frac{1}{x})}{h}\)

\(\displaystyle \frac {h}{h} = 1\)

final answer
\(\displaystyle 1\)

I don't know - what you are trying to do - it is so confused it is not even wrong!

\(\displaystyle f(x) \, = \, \frac{1}{x}\)

\(\displaystyle f(x+h) \, = \, \frac{1}{x+h}\)

\(\displaystyle \frac{f(x+h)-f(x)}{h} \, = \, \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \, = \, \frac{x - (x+h)}{h\cdot x \cdot (x+h)}\) <------Shouldn't the multiplication of \(\displaystyle x(x+h)\) to cancel out the denominator be on numerator part only and exclude the \(\displaystyle h\) in the denominator?

Now finish it ... carefully

\(\displaystyle \frac{x-x-h}{x^2h+xh^2}\)


final answer
\(\displaystyle -\frac {1}{x^2+xh}\)

 

[Deleted member 4993]

\(\displaystyle f(x) \, = \, \frac{1}{x}\)

\(\displaystyle f(x+h) \, = \, \frac{1}{x+h}\)

\(\displaystyle \frac{f(x+h)-f(x)}{h} \, = \, \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \, = \, \frac{x - (x+h)}{h\cdot x \cdot (x+h)}\)

\(\displaystyle \frac{x-x-h}{(x^2h+xh^2}\)

final answer
\(\displaystyle -\frac {1}{x^2+xh}\) <<<< Correct