Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 Re: Help Me ! » Discontinuous Derivative » 2010-02-10 08:27:27

How can you say f(0)=0? zero is not in the domain of  f. As a result, 0 is not in the domain of f' either. It has no meaning to check Continuity of a function out of its domain.

#2 Help Me ! » Discontinuous Derivative » 2010-02-08 03:26:09

Magister_Asmodeus
Replies: 3

Is it possible for the derivative of a function to be discontinuous? Attention! The derivative must be defined at that point, at which the derivative is discontinuous. I would like to see an example (if you believe it is possible) or a proof showing it is impossible.

#3 Re: Help Me ! » Tangent to circle » 2009-06-20 00:24:45

hi! noobard set the easy way. Here is a way that you can use in most exercises of this type.
In order for the line to be tangent to the circle, the distance between the centre of the circle and the line must be equal to the radius. The formula that gives the distance between a point M(x0,y0) and a line Ax+By+C=0 is d=|Ax0+By0+C|/√ (A²+B²)
So by solving the equation d=R (where R:radius) you can find the values of k you want.

#4 Re: Help Me ! » cost problem » 2009-06-02 03:41:22

Hi there.

Another thing you could say is:
since your Quadratic function is f(x)=(100-10x)(40+5x) or
f(x)=-50x²+100x+4000 with a=-50, b=100, c=4000
you can say that the maximum point will be (x,y)=(-b/2a,-Δ/4a), that means x=1 and y=4050
($45 per ticket, $4050 total income)

Board footer

Powered by FluxBB