Where has my mind gone?

bobrossu

New member
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Aug 16, 2017
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29
A couple math problems

Here are a few questions I need help with:


If sinx+cosx=Acos(x-B) for all real x where A and B are positive real numbers fund the maximum value of A+B.


For this one I think that I have to check all possible angles that fit with the rules given in order to isolate the values of A and B


Solve the inequality ee^x-e>0 (It's a power to a power).

Not sure about this one.

If f(x)=log(x!-20),where x is a natural number, what is the value of f-1(2)?

I know that x! is the factorial (1*2*...*x)
I inversed f(x) into f-1(x!)=10x+20 (Not sure if i'm right, most likely not.)
Ended up with f-1(2!)=10x+20 {equals to 10x=-18}
 
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If sinx+cosx=Acos(x-B) for all real x where A and B are positive real numbers fund the maximum value of A+B.

For this one I think that I have to check all possible angles that fit with the rules given in order to isolate the values of A and B

I'll assume that by Acos(x - B), you mean A * cos(x - B) and not the common notation for arccosine (aka inverse cosine). If that's the case, then my first step would be to use the angle difference identity to expand cos(x - B):

sin(x) + cos(x) = A[cos(x) cos(B) + sin(x) sin(B)]

[1 - A][sin(x) + cos(x)] = A[cos(B) + sin(B)]

Where does this take you?

Solve the inequality ee^x-e>0 (It's a power to a power).

Not sure about this one.

I'd begin by adding e to both sides and then taking the logarithm:

eex > e

ln(eex) > ln(e)

Try continuing from here.

If f(x)=log(x!-20),where x is a natural number, what is the value of f-1(2)?

I know that x! is the factorial (1*2*...*x)
I inversed f(x) into f-1(x!)=10x+20 (Not sure if i'm right, most likely not.)
Ended up with f-1(2!)=10x+20 {equals to 10x=-18}

Hmm... I actually don't know how to tackle this one, sorry. Maybe another member here can help you out.
 
Here are a few questions I need help with:


If sinx+cosx=Acos(x-B) for all real x where A and B are positive real numbers fund the maximum value of A+B.


For this one I think that I have to check all possible angles that fit with the rules given in order to isolate the values of A and B


Solve the inequality ee^x-e>0 (It's a power to a power).

Not sure about this one.

If f(x)=log(x!-20),where x is a natural number, what is the value of f-1(2)?

I know that x! is the factorial (1*2*...*x)
I inversed f(x) into f-1(x!)=10x+20 (Not sure if i'm right, most likely not.)
Ended up with f-1(2!)=10x+20 {equals to 10x=-18}

sin(x) + cos(x) = √2 * [1/√2 * sin(x) + 1/√2 * cos(x)]

continue.........

ee^x-e>0 →

ee^x > e1

ex> 1 ........................continue.........
 
Last edited by a moderator:
sin(x) + cos(x) = √2 * [1/√2 * sin(x) + 1/√2 * cos(x)]

continue.........

ee^x-e>0 →

ee^x > e1

ex> 1 ........................continue.........


Is the e a variable? If it is do i exclude some positive answers for x since e can be negative?
 
Is the e a variable? If it is do i exclude some positive answers for x since e can be negative?

No. In this case, e is the mathematical constant such that \(\displaystyle e^{ln(x)} = x\). Another way of defining it is \(\displaystyle \displaystyle e = \sum^{\infty}_{n=0} \: \frac{1}{n!}\). This definition results in \(\displaystyle e = 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + ... \approx 2.71828\)
 
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