few function etc problems..HELP!!

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April 30th 2012, 02:39 AM
ladieee

few function etc problems..HELP!!

If you can solve these i am very grateful. I am very bad at math. I tried to translate them in English from my language. Really hope you can understand what i mean.

1. What is the minimum value of function x^2 - 2^x + 1

2. what x values make function f(x) = 4x2 – 2x – 2 positive values?

3.
define the point of line 3x + 5y = 7, which is closest to the origin.

4. define the equation of line, which is normal of line y = -3x + 2 and goes through the point (6, 8)

5. Solve the following equations using logarithms to two decimal places.

a)=8 x120
b) 4∙12x= 40
c) 7-2x -20 =0

thank you very very much!!!!:)
April 30th 2012, 03:08 AM
Prove It

Re: few function etc problems..HELP!!

Quote:

Originally Posted by ladieee

If you can solve these i am very grateful. I am very bad at math. I tried to translate them in English from my language. Really hope you can understand what i mean.

1. What is the minimum value of function x^2 - 2^x + 1

2. what x values make function f(x) = 4x2 – 2x – 2 positive values?

3.
define the point of line 3x + 5y = 7, which is closest to the origin.

4. define the equation of line, which is normal of line y = -3x + 2 and goes through the point (6, 8)

5. Solve the following equations using logarithms to two decimal places.

a)=8 x120
b) 4∙12x= 40
c) 7-2x -20 =0

thank you very very much!!!!:)

Is the first function $\displaystyle \begin{align*} x^2 - 2^x + 1 \end{align*}$ or $\displaystyle \begin{align*} x^2 - 2x + 1 \end{align*}$ ?
April 30th 2012, 03:23 AM
ladieee

Re: few function etc problems..HELP!!

Quote:

Originally Posted by Prove It

Is the first function $\displaystyle \begin{align*} x^2 - 2^x + 1 \end{align*}$ or $\displaystyle \begin{align*} x^2 - 2x + 1 \end{align*}$ ?

Its the first you mentioned:)
April 30th 2012, 03:30 AM
Plato

Re: few function etc problems..HELP!!

Quote:

Originally Posted by ladieee

Its the first you mentioned:)

If $y=x^2+2^x+1$ then $y'=2x+2^x\cdot \ln(2)$ .
Now solve $y'=0.$
April 30th 2012, 03:39 AM
Prove It

Re: few function etc problems..HELP!!

Quote:

Originally Posted by ladieee

Its the first you mentioned:)

Minima are found at points where the derivative is 0 and where the second derivative is positive.
April 30th 2012, 07:01 PM
HallsofIvy

Re: few function etc problems..HELP!!

Quote:

Originally Posted by ladieee

If you can solve these i am very grateful. I am very bad at math. I tried to translate them in English from my language. Really hope you can understand what i mean.

1. What is the minimum value of function x^2 - 2^x + 1

f'= 2x+ 2^x ln(2)= 0 can not be solved in terms of elementary functions. It can be solved numerically or using the Lambert W function.

Quote:

2. what x values make function f(x) = 4x2 – 2x – 2 positive values?

$4x^2- 2x- 2= 2(2x^2- x- 1)= 2(2x+1)(x- 1)$ . That will be positive when all three factors are positive or when 2x+1 and x- 1 are both are negative. That is, solve (2x+1> 0 and x- 1> 0) or (2x+1< 0 and x- 1< 0).

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3.

Quote:

define the point of line 3x + 5y = 7, which is closest to the origin.

5y= 7- 3x so y= (7/5)- (3/5)x. You could write $d(x,y)= \sqrt{x^2+ y^2}= \sqrt{x^2+ ((7/5)- (3/5)x)^2}$ . Differentiate this with respect to x, set it equal to 0, and solve for x. It is a little simpler to note that x has a max or min if and only if $x^2$ has a max or min so you could just look at $f(x)= x^2+ ((7/5)- (3/5)x)^2$ .

But the simplest way to approach this is geometrically. The shortest distance from a point to a line is always along the line, through the point, perpendicular to the given line. The slope of the line 3x+ 5y= 7 is -3/5 so any line perpendicular to it has slope 5/3. The line through the origin, with slope 5/3, is y= (5/3)x or 5x- 3y= 0. Determine where that line crosses the line 3x+ 5y= 7 and find the distance from the origin to that point.

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4. define the equation of line, which is normal of line y = -3x + 2 and goes through the point (6, 8)

Did you even try to do this yourself? that line has slope -3 so any line normal to it has slope 1/3. The line through (6, 8) with slope 1/3 is y= (1/3)(x- 6)+ 8.

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5. Solve the following equations using logarithms to two decimal places.

a)=8 x120

It's almost impossible to tell what you mean. PLEASE use "^" to indicate an exponent. And the "=" doesn't make sense where it is. I think you mean 8^x= 120. If so, just take the logarithm of both sides and apply the laws of logarithms: log(8^x)= x log(8)= 120.

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b) 4∙12x= 40

If you mean 4(12^x)= 40 then start by dividing both sides by 4: 12^x= 10 so log(12^x)= xlog(12)= log(10).

Quote:

c) 7-2x -20 =0

This one I am not at all sure about. The best I can make of it is 7- 2^x- 20= 0 which is the same as 2^x= 14.
log(2^x)= x log(2)= log(14).

Quote:

thank you very very much!!!!:)

I'm not sure I have done you any favor. You say you are "bad at math". The way you become good at math is by doing math, not by having some one else do it for you. I hope I have left enough for you to do to help you become better.