rationalize this numerator so that
[(√2(x+delta x)-3) - (√2x-3)]/delta x = 2/[(√2(x+delta x)-3) + √2x-3)]
Hello, yess!
Is there a typo on the problem?
$\displaystyle \text{Rationalize this numerator so that:}$
.$\displaystyle \displaystyle \frac{\sqrt{2}(x+ \Delta x)-3) - (\sqrt{2}x-3)}{\Delta x} \;=\; \frac{2}{[\sqrt{2}(x+\Delta x)-3] + [\sqrt{2}x-3]}$
This is not true . . .