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