So I have to find the derivative of f(x) = |x-2|, but I don't know how to get there algebraically when I evaluate the derivative for the limit h --> 0 of ((f(x+h) - f(x))/ h))
I have to evaluate the function as the limit h approaches 0, and I know the function will be piecewise because of the absolute value, but I need to be able to use this method and show all of the steps that I used. Detailed answers would be ideal, thank you! I do not earn any points for this, but I need to know how to do it.
I have to evaluate the function as the limit h approaches 0, and I know the function will be piecewise because of the absolute value, but I need to be able to use this method and show all of the steps that I used. Detailed answers would be ideal, thank you! I do not earn any points for this, but I need to know how to do it.
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f(x) = x - 2 (for x > 0)
f(x + h) = x + h - 2
f(x + h) - f(x) = x + h - 2 - (x - 2) = h
[f(x + h) - f(x)] / h = 1
f ' (x) = 1 for x > 0
f(x) = 2 - x (for x < 0)
f(x + h) = 2 - (x + h) = 2 - x - h
f(x + h) - f(x) = 2- x - h - (2 - x) = -h
[f(x + h) - f(x)] / h = -1
f ' (x) = -1 for x < 0
f ' (x) is undefined at x = 0 because the LH and RH limits of slope will be different
f(x + h) = x + h - 2
f(x + h) - f(x) = x + h - 2 - (x - 2) = h
[f(x + h) - f(x)] / h = 1
f ' (x) = 1 for x > 0
f(x) = 2 - x (for x < 0)
f(x + h) = 2 - (x + h) = 2 - x - h
f(x + h) - f(x) = 2- x - h - (2 - x) = -h
[f(x + h) - f(x)] / h = -1
f ' (x) = -1 for x < 0
f ' (x) is undefined at x = 0 because the LH and RH limits of slope will be different
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oops, you're right...
f ' (x) = 1 for x > 2
f ' (x) = -1 for x < 2
f ' (x) undefined for x = 2 because the LH and RH limits of slope are different
f ' (x) = 1 for x > 2
f ' (x) = -1 for x < 2
f ' (x) undefined for x = 2 because the LH and RH limits of slope are different
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fx = |x - 2|=> for real x values you can re-write as:
= √(x - 2)^2 => now go ahead and do your stuff.
= √(x - 2)^2 => now go ahead and do your stuff.