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1. Find the greatest integer x such that \(|x^2 - 98x+1| = 98x-x^2-1.\)

 

2. Find the greatest possible value for \( \frac{\sqrt{\lfloor x \rfloor}}{\lfloor \sqrt{x} \rfloor},\)where \(x \ge 1. \)

 

3. Let \\(f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.\)Find a+b if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper).

 

Thank you for the help!

 Jun 20, 2019
 #1
avatar+101813 
+1

3.

 

We have that

 

2(-2)- b  = -2 - 5             and           a(2) + 3  = 2 - 5      

-4- b   = -7                                        2a + 3  = -3

b = 3                                                 2a  = -6

                                                           a = -3

 

Here's the graph : https://www.desmos.com/calculator/avrddz9mrw

 

And a + b  =   0

 

 

cool cool cool

 Jun 20, 2019
 #5
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0

Thank you !

Guest Jun 21, 2019
 #2
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0

xxxxxxxxxxxxx.

 Jun 20, 2019
edited by Guest  Jun 20, 2019
 #3
avatar+22527 
+2

1.
Find the greatest integer x such that \(|x^2 - 98x+1| = 98x-x^2-1\).

 

\(\begin{array}{|rclrll|} \hline \mathbf{|x^2 - 98x+1|} &=& \mathbf{98x-x^2-1} \\\\ |x^2 - 98x+1| &=& -\underbrace{(x^2 - 98x+1)}_{<0!} \\\\ && x^2 - 98x+1 &<& 0 \\ && \left(x-\dfrac{98}{2}\right)^2 -\dfrac{98^2}{4} + 1 &<& 0 \\ && \left(x-\dfrac{98}{2}\right)^2 &<& \dfrac{98^2}{4} - 1 \\ && \left(x-\dfrac{98}{2}\right)^2 &<& \dfrac{98^2-4}{4} \\ && x-\dfrac{98}{2} &<& \dfrac{\sqrt{98^2-4}}{2} \\ && x &<& \dfrac{98}{2} +\dfrac{\sqrt{98^2-4}}{2} \\ && x &<& \dfrac{98+\sqrt{98^2-4}}{2} \\ && x &<& 97.9897948557 \\ &&\mathbf{ x_{\text{max}}} &=& \mathbf{97} \\ \hline \end{array} \)

 

laugh

 Jun 20, 2019
 #6
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+1

Thank you so much!

Guest Jun 21, 2019
 #4
avatar+28064 
+3

This should help answer question 2:

 

 Jun 20, 2019
 #7
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0

Thank you!

Guest Jun 21, 2019

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