Y yess New member Joined Sep 17, 2010 Messages 5 Sep 23, 2010 #1 rationalize this numerator so that [(?2(x+delta x)-3) - (?2x-3)]/delta x = 2/[(?2(x+delta x)-3) + ?2x-3)]
rationalize this numerator so that [(?2(x+delta x)-3) - (?2x-3)]/delta x = 2/[(?2(x+delta x)-3) + ?2x-3)]
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Sep 24, 2010 #2 Hello, yess! Where did this problem come from? There must be a typo . . . \(\displaystyle \text{Rationalize this numerator so that:}\) . . . . \(\displaystyle \frac{[\sqrt{2}(x+\Delta x)-3] - [\sqrt{2}x-3]} {\Delta x} \;=\; \frac{2}{[\sqrt{2}(x+\Delta x)-3] + [\sqrt{2}x-3]}\) Click to expand... This is not true . . . The left side is the Difference Quotient for: .\(\displaystyle f(x) \:=\:\sqrt{2}\,x - 3\) I can't imagine why anyone would want to rationalize the numerator. . . (And get the wrong result.)
Hello, yess! Where did this problem come from? There must be a typo . . . \(\displaystyle \text{Rationalize this numerator so that:}\) . . . . \(\displaystyle \frac{[\sqrt{2}(x+\Delta x)-3] - [\sqrt{2}x-3]} {\Delta x} \;=\; \frac{2}{[\sqrt{2}(x+\Delta x)-3] + [\sqrt{2}x-3]}\) Click to expand... This is not true . . . The left side is the Difference Quotient for: .\(\displaystyle f(x) \:=\:\sqrt{2}\,x - 3\) I can't imagine why anyone would want to rationalize the numerator. . . (And get the wrong result.)
