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OMG THANK YOU SO MUCH. Thx. 

x^k

If k >  0

Then the range  on  [1, infinity)  will be    [1, 0)

Here's some graphs for selected values of k......note that they all have the same range on  [1, inf)

https://www.desmos.com/calculator/eldeljuf1q

\( x\in (-\infty, 1)\cup(5,\infty) \)

-x^2 + bx - 5 <  0    (1)      multiply though by  - 1   and  change the  direction of the  inequalitiy sign

x^2 - bx  + 5  >   0        set = to  0   

x^2 - bx + 5   =  0

We need to  have these  factors

(x - 1)  ( x - 5)     expanding we  get that

x^2  - 6x  +  5   =  0    so....this implies that we actually need

x^2 - 6x + 5   >  0        (2)

(x - 1) ( x - 5)  >  0

Note that  when    x =  ( -inf, 1)     both linear factors will be  negative.....so....their  product will be  positive

And  when x = ( 5, inf)   the same will occur

So  multiplying (2)  back  through by  -1   will produce  (1)   and we have that

-x^2 + 6x - 5 <  0 ....so......b  = 6

Here's a graph to show that this function is < 0   on  (-inf, 1)   and (5, inf)

https://www.desmos.com/calculator/uwy0jmousq

2) Let  z  =  a + bi

So we have

2 (a + bi)  - 3 (a - bi)   =  -2  - 30i      simplify

2a + 2bi   - 3a  + 3bi  =   -2  - 30i

-1a + 5bi =   -2 - 30i

Equate terms

-1a   = -2             5bi  = -30i

a  =  2           b  = -6

So

z =  2  -  6i   

I just got an answer here is the link 

https://web2.0calc.com/questions/math-help-srry-for-inconviance

Hope this helps

(a) Is it possible for x=3 to be a root of f(x)?
 

Yes.....suppose we  have 

(x - 1)^2 ( x - 2) ( x - 3)......x = 3 is a root.....and expanding this we  get......x^4 - 7 x^3 + 17 x^2 - 17 x + 6

(b) Is it possible for x = 3 to be a double root of f(x)?

Let's suppose that it is possible

Let the other roots be   a, b     where a, b are integers....and we have that

(x - a) ( x - b) ( x - 3) ( x - 3)  =

(x^2 - (a + b) x + ab) ( x^2 - 6x + 9)  =

x^4 -(a + b)x^3 + abx^2  - 6x^3 + 6(a+b)x^2 -6abx + 9x^2 - 9(a + b)x + 9ab

This implies that    9ab   = 6  ⇒    ab  = 6 / 9  ⇒     ab  =  2 / 3

But a,b are integers.....and  integers multiplied together  must  produce an integer......

So...it is not possible  fro x = 3 to be a  double-root in this case

Bad day at the office Rom ?

Let \(x,y,z \) be nonnegative real numbers such that \(x + y + z = 3\)

Find the maximum value of \((xy + z)(xz + y)\)

\(\begin{array}{|rcll|} \hline \huge{GM} &\huge{\leq}& \huge{AM} \\\\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{(xy + z)+(xz + y)}{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ xy + z + xz + y }{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{x(y+z)+(y+z)}{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (y+z)(x+1)}{2} \quad | \quad \boxed{y + z = 3-x} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (3-x)(x+1)}{2} \\ &&\begin{array}{|rcll|} \hline \large{GM} &\large{\leq}& \large{AM} \\\\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ (3-x)+(x+1)}{2} \\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ 3-x + x+1 }{2} \\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ 4 }{2} \\ \sqrt{(3-x)(x+1)} & \leq & 2 \quad | \quad \text{square both sides}\\ (3-x)(x+1) & \leq & 4 \\ \mathbf{ \dfrac{(3-x)(x+1)}{2}} & \leq & \mathbf{ 2 } \\ \hline \end{array} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (3-x)(x+1)}{2} \leq 2 \quad | \quad \text{ The maximum value results from the equals sign} \\ \sqrt{(xy + z)(xz + y)} & = & \dfrac{ (3-x)(x+1)}{2} = 2 \quad | \quad \text{square both sides}\\ \mathbf{(xy + z)(xz + y) } &=& \mathbf{4} \\ \hline \end{array}\)

The maximum value of \(\mathbf{(xy + z)(xz + y)=4}\)

Here are the three examples of inline LaTeX: Rendered

\({}\)

Standard writing in ascii followed by LaTeX equation:  \(4*(\frac{1}{2}x)+x=432\); followed by more ascii text.

Standard writing in ascii followed by LaTeX equation (inside parentheses) in LaTeX:  \((4*(\frac{1}{2}x)+x=432)\);  followed by more ascii text.

Standard writing in ascii followed by LaTeX equation (inside parentheses) with text in LaTeX:  \((4*(\frac{1}{2}x)+x=432) \text { text in LaTeX}\);  followed by more ascii text.

One more example where the rendered LaTeX is offset and centered.

Test for inline use of LaTeX \[\text {Quadratic Formula } x = {-b \pm \sqrt{b^2-4ac} \over 2a}\] End of test.

To do this, use \[ \] (squared brackets in place of the parentheses) 

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Edited added additional example, improved appearance.  

GA

^{(Updated and corrected post)}

Yes, there is a delimited code sequence to activate LaTeX for inline presentations.

The sequence is “\(” (backslash -open parenthesis –no quotes) to start the LaTex  and  “\)”  (backslash -close parenthesis –no quotes) to terminate and return to ascii mode.

There is one additional requirement to confirm the rendering the LaTeX:

Somewhere (anywhere) in the post must be this (compiler) command:

<span class="math-tex">\({}\) span>  

Or

Open a LaTeX box and type  {} < --- this is the easy way.

Note that these commands are functionally equivalent. They use a null character, becoming (and remaining) invisible after the first preview when returning to edit mode. 

Additional Notes:

 If the post already contains one or more standard LaTeX boxes then this is not necessary.  Without this command, the display will not render the LaTeX code. With this command, it always works, if the system is functioning normally.   

Here are three examples of inline LaTeX presented unrendered, in ASCII format:

Standard writing in ascii followed by LaTeX equation:  \(4*(\frac{1}{2}x)+x=432 \); followed by more ascii text.

Standard writing in ascii followed by LaTeX equation (inside parentheses) in LaTeX:  \((4*(\frac{1}{2}x)+x=432)\);  followed by more ascii text.

Standard writing in ascii followed by LaTeX equation (inside parentheses) with text in LaTeX:  \((4*(\frac{1}{2}x)+x=432) \text { text in LaTeX }\);  followed by more ascii text.

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The examples coded in this post have hidden modifications to prevent rendering the LaTeX. The examples show just the ascii code for the LaTeX.  The post below does not have these modifications, so it will render the LaTeX in inline mode.

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 ^{Edits: Error correction, improved readability and coherency, and modified the examples to display the exact LaTeX code instead of rendering.}

GA

Compute  \(\Bigg\lfloor \dfrac{2007! + 2004!}{2006! + 2005!}\Bigg\rfloor\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \Bigg\lfloor \dfrac{2007! + 2004!}{2006! + 2005!}\Bigg \rfloor }\\ &=& \Bigg\lfloor \dfrac{2007!}{2006! + 2005!} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006!~2007}{2005!~(1+2006)} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006!~2007}{2005!~2007} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006! }{2005! } + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2005!~2006 }{2005! } + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor 2006 + \underbrace{\dfrac{2004!}{2006! + 2005!}}_{<1} \Bigg \rfloor \\ &=& \mathbf{2006} \\ \hline \end{array}\)

you have to start with k=1.

The answer is 12

\(f(2) = 5(2^2) - \dfrac 1 2 + 3 = \dfrac{45}{2}\\ g(2) = 2^2 - k = 4-k\\ f(2)-g(2)=2\\ \dfrac{45}{2}-(4-k) = 2\\ \dfrac{37}{2} + k = 2\\ k = -\dfrac{33}{2}\)

\(h \otimes h = h^3-h\\ h \otimes (h\otimes h) = \\ h \otimes (h^3-h)=\\ h^3 - (h^3-h) =\\ h\\~\\ h\otimes (h \otimes h) = h\)

\(\text{I'm assuming the braces stand for the fractional part of a number. $\{x\}=x-\lfloor x \rfloor$}\\ \{3\} = \sqrt{3}-1\\ \{2\}=\sqrt{2}-1\\~\\ z = \dfrac{(\sqrt{3}-1)^2-2(\sqrt{2}-1)^2}{(\sqrt{3}-1)-2(\sqrt{2}-1)}\)

\(z = \dfrac{(3-2\sqrt{3}+1)-2(2-2\sqrt{2}+1)}{\sqrt{3}-2\sqrt{2}+1}=\\ \dfrac{-2-2\sqrt{3}+4\sqrt{2}}{\sqrt{3}-2\sqrt{2}+1} = \\ -2\cdot \dfrac{\sqrt{3}-2\sqrt{2}+1}{\sqrt{3}-2\sqrt{2}+1} = -2\)

\(\lfloor -2 \rfloor = -2\)

\(\text{We need to ensure that there is no discontinuity at $x=4$}\\ f(4) = 5\\ \lim \limits_{x\to 4^-} f(x) = 4+k\\ 5 = 4+k\\ k=1\)

\(\left \lfloor \dfrac{2007! + 2004!}{2006!+2005!}\right\rfloor = \\ \left \lfloor \dfrac{2004!(1+2005\cdot 2006 \cdot 2007)}{2004!\cdot 2005(1+2006)}\right\rfloor = \\ \left \lfloor \dfrac{1+2005\cdot 2006 \cdot 2007}{2005(1+2006)}\right\rfloor = \\ \left \lfloor \dfrac{\frac{1}{2005}+2006 \cdot 2007}{1+2006}\right\rfloor = \\ \left \lfloor \dfrac{\frac{1}{2005\cdot 2006}+2007}{\frac{1}{2006}+1}\right\rfloor = 2006\\\)

The last equality comes from the first term in the numerator being so small we can ignore it, 

and the denominator being slightly greater than 1 so that the floor is 2006 rather than 2007.

\(\left|\dfrac 1 2 - c i \right| = \left(\dfrac 1 2 \right)^2 + c^2 = \dfrac 3 4\\ \dfrac 1 4 + c^2 = \dfrac 3 4\\ c^2 = \dfrac 1 2\\ c = \pm \dfrac{1}{\sqrt{2}}\)

\(2z - 3\bar{z} = -2 - 30i\\ \text{let $z=x+iy$}\\ 2x + i2y - 3x + i3y = -2 - 30i\\ -x+i5y=-2-30i\\ x = 2,~y = -6i\\ z = 2-6i\)

Let P be the original price of the  stock

So we have, after  two  days

[ P (.80) ] (1.30)   = 

P  * 1.04  ⇒      i.e., the  stock  has increased in value by 4%

Two is a prime number and it’s even so false

Isn't it 6 integers as follows:

2, 3, 4, 5, 6, 7.

1/2, 1/3, 1/4, 1/5, 1/6 and 1/7 - which gives 3 terminating and 3 non-terminating

So, the smallest positive integer n = 7

Wouldn't the final quadratic be \(\frac{a}{b}(x^2+3x-10)\) where the entire quadratic is multiplied by a/b, instead of just x^2?