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!!!!:)

Re: few function etc problems..HELP!!

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

Re: few function etc problems..HELP!!

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

Its the first you mentioned:)

Re: few function etc problems..HELP!!

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Originally Posted by

**ladieee** Its the first you mentioned:)

If $\displaystyle y=x^2+2^x+1$ then $\displaystyle y'=2x+2^x\cdot \ln(2)$.

Now solve $\displaystyle y'=0.$

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.

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.

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

$\displaystyle 4x^2- 2x- 2= 2(2x^2- x- 1)= 2(2x+1)(x- 1)$. 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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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 $\displaystyle d(x,y)= \sqrt{x^2+ y^2}= \sqrt{x^2+ ((7/5)- (3/5)x)^2}$. 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 $\displaystyle x^2$ has a max or min so you could just look at $\displaystyle f(x)= x^2+ ((7/5)- (3/5)x)^2$.

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.

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).

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).

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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.