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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