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# 1. Find the greatest integer x such that

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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
+104704
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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

Jun 20, 2019
#5
0

Thank you !

Guest Jun 21, 2019
#2
0

xxxxxxxxxxxxx.

Jun 20, 2019
edited by Guest  Jun 20, 2019
#3
+23277
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1.
Find the greatest integer x such that $$|x^2 - 98x+1| = 98x-x^2-1$$.

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

Jun 20, 2019
#6
+1

Thank you so much!

Guest Jun 21, 2019
#4
+28179
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This should help answer question 2:

Jun 20, 2019
#7
0

Thank you!

Guest Jun 21, 2019