This is a question on my summer math packet for Calculus AB AP. I haven't learned anything about derivatives yet but when I tried to google this question to find a way to approach it almost all the hits involved derivatives, which I do not know how to do yet.
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f(x) = x² – 2x – 3
Now write (x + Δx) in place of x
f(x + Δx) = (x + Δx)² – 2(x + Δx) – 3
Now write (x + Δx) in place of x
f(x + Δx) = (x + Δx)² – 2(x + Δx) – 3
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It is straightforward, you just have to replace x by x+h in the expression of you function (h means delta x here, which is nothing but a notation)...
f(x+h)= (x+h)^2 - 2(x+h) - 3 = (x^2 +h^2 + 2hx) - 2x-2h -3 = x^2 +x(2h-2) + h^2 -2h -3
With this expression you see that f(x+h) = f(x) + h^2 - 2h + 2xh
And then if you need (f(x+h)-f(x))/h you will find h-2+2x
Now if you take h as small as possible (in absolute value), which means h tends to zero then you will find the derivative function of your function : namely g(x)= 2x-2
Indeed the definition of the derivative function (if it exists in a given interval) of a function f, is the limit when h tends to zero of the following expression: (f(x+h)-f(x))/h
f(x+h)= (x+h)^2 - 2(x+h) - 3 = (x^2 +h^2 + 2hx) - 2x-2h -3 = x^2 +x(2h-2) + h^2 -2h -3
With this expression you see that f(x+h) = f(x) + h^2 - 2h + 2xh
And then if you need (f(x+h)-f(x))/h you will find h-2+2x
Now if you take h as small as possible (in absolute value), which means h tends to zero then you will find the derivative function of your function : namely g(x)= 2x-2
Indeed the definition of the derivative function (if it exists in a given interval) of a function f, is the limit when h tends to zero of the following expression: (f(x+h)-f(x))/h
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If it said "write an expression for f(x + y) if f(x) = x^2 - 2x -3", would you know what to do? Do that!
Now imagine that "delta x" is just another way of writing "y". You can now write out an expression for "f(x + delta x)"
Now imagine that "delta x" is just another way of writing "y". You can now write out an expression for "f(x + delta x)"