# Help (Challenge proplems)

1. Jun 22, 2007

### majdeee

I tried this for the first one but

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2. Jun 22, 2007

### bel

for the 1st question, let b=x, for the sake of tradition.

find d/dx (e^x-x^e)

find d/dx (e^x-x^e)=0 on the interval [0, infinity)

prove that the zero was a minium, substitute the x it into y= e^x-x^e

if the result is y=0, you have proved it

thinking about the second question and the third question.

Last edited: Jun 22, 2007
3. Jun 22, 2007

### Office_Shredder

Staff Emeritus
To find all the zeroes of that derivative, you probably essentially need to do the original problem anyway

4. Jun 22, 2007

### bel

nope, graph it and see

5. Jun 22, 2007

### PatF

1) For the first problem, look at the function f(x)=log(x)/x when x>=1.

You can easily prove that it has a maximum at e. But then,

log(e)/e>=log(b)/b

whenever b>=1. This says b>=e*log(b). If you exponentiate this, you find

e^b>=b^e

2) For the second problem, look at the function f(x)=x^x. You can easily how that this has a minimum at 1/e. However, the limit as x goes to zero of x^x is 1. Furthermore, 1^1 is also 1. Thus x^x decreases between 0 and 1/e and increases between 1/e and 1. So if you have a number, say a, between 0 and 1/e, there must be a number between 1/e and 1, say b such that a^a=b^b.

HTH.

--PatF

6. Jun 22, 2007

### Staff: Mentor

Thread moved to Homework Help forums.

7. Jun 23, 2007

### majdeee

this is the answer for the first 1
we need to prove e^b>=b^e
since ln(x) (natural lorarithm of x is continuous and monotonic increasing function we can take the natural logarithm of both sides of the inequality without disrupting it
so we may prove ln(e^b)>=ln(b^e)
by simplification
ln(e^b)=b*ln(e)=b*1=b
ln(b^e)=e*ln(b)
so we only have to prove that b>=e*ln(b)
it is readiely seen true for b=1 and b=e
(for b=e) we have equality
let's look at the two sides of the inequality as two functions of b
f1(b)=b
f2(b)=e*ln(b)
the derivatives of the functions are
df1(b)/db=1
df2(b)/db=e/b
for b>e df1(b)/db=1 and df2(b)/db=e/b<1
so that from value of b>e f1 increases faster than f2 and since they both start at the same value at b=e the inequality holds for b>e
for 1<b<e we have df2(b)/db=e/b>1 since b<e examining the regression of b from e to 1 it is regressing at a rate that is larger than 1 after starting from equal value of e it goes down faster than b therefore we have proven

8. Jun 23, 2007

### majdeee

does anybody know how to solve the third one??:surprised

9. Jun 23, 2007

### malawi_glenn

For the third one, I would beging with:

P(x) = Ax^2 + Bx + C ; A,B,C are rational coefficients.

Then I would a partial fraction decomposition:

$$R(x) = \dfrac{Ax^{2} + Bx + C}{x^{3} (x-1)} = \dfrac{a_{0} ^{(1)}}{x^{3}} + \dfrac{a_{0} ^{(2)}}{x^{2}} + \dfrac{a_{0} ^{(3)}}{x} + \dfrac{a_{1} ^{(1)}}{x-1}$$

To find out the coefficients $$a_{1} ^{(1)}$$....

; you just write

$$R(x) = \dfrac{P(x)}{x^{3} (x-1)}$$

then you get a nice linear system of equations.

After this, do the integration (remember the constant of integration), then see what values of $$a_{1} ^{(1)}$$ , and hence also values of A,B and C, that gives you a rational function.