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The term "f of x" is used in calculus to indicate some undefined function that is being applied to the variable x; normally this would appear in the form of an equation, which would then tell us what this function does to x.

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Q: What is F of X?
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What is an antidifference?

In mathematics, a function F(x) is the antidifference of f(x) if F(x+1)-F(x)=f(x).


When was Pierre De fermat's last theorem created?

PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.


What is the integral of f divided by the quantity 1 minus f with respect to x where f is a function of x?

∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx


Formula for instantaneous speed?

lim as h->0 of (f(x+h) - f(x))/h or lim as x->a of (f(x) - f(a))/(x - a)


What is the first derivative used for?

The first derivative f'(x) gives the instantaneous slope of f(x). If f'(x) is positive, then f(x) is increasing (positive slope), and if f'(x) is negative, then f(x) is decreasing (negative slope). If f'(x) = 0, then the graph of f(x) is flat at the point (slope = 0).