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F(x)=x^2-3x find (F(x+h)-F(x))/h
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the difference quotient beloved by math teachers
f(x+h) = (x+h)^2 -3(x +h) = x^2 +2xh +h^2 -3x -3h f(x) = x^2 - 3x f(x+h) - f(x) = x^2 +2xh +h^2 -3x -3h -x^2 +3x = h^2 - 3h (F(x+h)-F(x))/h = (h^2 -3h)/ h = h(h-3) / h = h-3
only hard part is figuring out what \[f(x+h)\] means in this case it is \[f(x+h)=(x+h)^2-3(x+h)\] \[=x^2+2xh+h^2-3x-3x\]
typo \[=x^2+2xh+h^2-3x-3h\] so \[f(x+h)-f(x)=x^2+2xh+h^2-3x-3h-(x^2-3x\] \[=x^2+2xh+h^2-3x-3h-x^2+3x\] \[=2xh+h^2-3x\]
another typo \[=2xh+h^2-3h\]
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as usual everything without an h in it has gone. so finally \[\frac{f(x+h)-f(x)}{h}=\frac{2xh-3h+h^2}{h}=\frac{h(2x-3+h)}{h}=2x-3+h\]
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