Math 1507

October 16, 1997
Tom Morley
Name: ____________________________

Open Book and Notes. You have one half hour. Please explain all answers.

Problem 1

Find all x such that | 12 - 4x | ⩾ | x + 2|

Solve[Abs[12  -  4 x]  ==  Abs[x  +  2], x]

So to draw a picture

Plot[  Abs[ 12 - 4 x] - Abs[x + 2],{x,-1,10}]

Check the signs in the various intervals. We condude that the answer is:


                                                       2 ⩾ x  or  x ⩾ (14/3)

Note: While I expect that this is the method that you will use "Out in the real world", I am expecting an anwer on the test that examines the 4 cases:

I.   12 - 4x >0 and x + 2>0,
II. 12 - 4x >0 and x + 2 <0
III. 12 - 4x<0 and x + 2>0
IV. 12 - 4x <0 and x+2 <0

Problem  2

Let duffin(x) be the following function. (Actually is the the 
third order Bessel function of the second kind.)

Plot[Duffin[x],{x,1,10}]

It turns our that duffin'(x) is

   erdos((x^2)), where erdos is given in terms a sum of certain modified bessel funcitons of the second kind.

Differentiate with respect to x:

sin(1/((x^3) + 3 (x^2)) )

D[Sin[1/(x^3 + 3 x^2)],x]

Clear[duffin]

duffin' = erdos[#^2]&

D[duffin[x^2],x]

duffin((x^3)) sin(x)

D[duffin[x^3] Sin[x],x]

duffin(x)/((x^3) + 3 (x^2))

D[duffin[x]/(x^3 + 3 x^2),x]

Problem 3

Find the left and right limits as x -> 9 of    :f(x) =    ⌊-  (Sqrt[x]) ⌋ x . Explain and justify your andwer.

f[x_] := Floor[-Sqrt[x]] x

Plot[f[x],{x,8,10}]

If x is nearby 9, but slightly smaller, then   x < 9, and thus (Sqrt[x])  < 3 (but not by much).  Therefore - (Sqrt[x])  > -  3 (BNBM), and thus ⌊ -(Sqrt[x])  ⌋ = -3 provided that x < 9, but not by much. Therefore if  x < 9, but not by much,   ⌊ -(Sqrt[x])  ⌋  x = -3x, and thus the left limit as 
x -> 9 of  ⌊ -(Sqrt[x])  ⌋  x  is the same as the left limit as x -> 9 of - 3x. This limit is -27

If x is nearby 9, but slightly bigger, x > 9. Tthus (Sqrt[x])  > 3 (but not by much), therefore 
- (Sqrt[x])  < -  3 (BNBM), and thus ⌊ -(Sqrt[x])  ⌋ = -4 provided that x > 9, but not by much. Therefore if  x > 9, but not by much,   ⌊ -(Sqrt[x])  ⌋  x = -4x, and thus the right limit as
 x -> 9 of  ⌊ -(Sqrt[x])  ⌋  x  is the same as the left limit as x -> 9 of - 4x. This limit is -36.

Problem 4

Differentiate by sight.Explain your answers and be sure
to indicate  appoximate scale.

Duffin[x_] = BesselY[3,x]

Plot[Duffin[x],{x,1,12}]

Plot[(x-2)^2 (x+1) (x - 4), {x,-2,5}]

Plot[BesselJ[2,x^2],{x,-1,3}, PlotRange-> All]

Answers:

Plot[Duffin'[x],{x,1,12}]

Plot[Evaluate[D[(x-2)^2 (x+1) (x - 4),x]], {x,-2,5}]

Plot[Evaluate[D[
     BesselJ[2,x^2],x]],{x,-1,3}, PlotRange-> All]