evaluate (f(x+h)-f(x))/h and simplify if f(x)=x^2-2x
2x -2 :)
shortcut ? :)
it gives you the instruction; are you stuck someplace? or just want someone to do it for you?
f(x+h)=(x+h)^2-2(x+h)
evaluate using the above. how far did you get ?
then only h104-28ard part of this is understanding what \[f(x+h)\] means. it means you have to replace x in f(x) by x +h you get \[f(x+h)=(x+h)^2-2(x+h)\] so \[f(x+h)-f(x)=(x+h)^2-2(x+h)-(x^2-2x)\] and \[\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2-2(x+h)-(x^2-2x)}{h}\]
its commie code, i just know it
h104-28ard=hard ?
ah pellets and retrices lol
this is the "difference quotient so beloved by pre-calc teachers to get you ready for calculus. it represents the change in y over the change in x like the slope. now you do a raft of algebra and a miracle will occur, everything without h will disappear from the numerator
yeah you broke the code! pellet
h might still remain above; but it will vanish from below; and when it goes to 0; we are left with the 2x-2 part
numerator is \[x^2+2xh+h^2-2x-2h-x^2+2x=2xh-2h+h^2\] so you have \[\frac{2xh-2h+h^2}{h}=\frac{h(2x-2+h)}{h}=2x-2+h\]
@amistre you are jumping the gun. didn't get to the "let h go to zero" part yet i bet
♫♫♫ bang bang bang goes the trolley ♫♫♫
it is still hanging around like a distance cousin after thanksgiving
ok that makes a lot more sense now Thank you!!
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