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1) Consider the given functions:

\(\begin{array}{ccc} f(x) & = & 5x^2 - \frac{1}{x}+ 3\\ g(x) & = & x^2-k \end{array}\)

 

If\( f(2) - g(2) = 2\), what is the value of \(k\)?

 

2) What is the greatest integer value of \(b\) such that\(-4 \) is not in the range of \( y=x^2+bx+12\)?

 

3)For any number \(x\), we are told that \(x\&=7-x\) and \(\&x = x -7\). What is the value of \(\&(12\&)\)?

 

4)Let \(f(x) = 3x + 3\) and \(g(x) = 4x + 3.\) What is \(f(g(f(2)))\)?

 

THX IN ADVANCE

 Aug 16, 2019
 #1
avatar+106536 
+1

(1)

 

5(2)^2  - 1/2  + 3   -  ( 2^2 - k)   =    2

 

20  - 1/2 + 3   - 4  + k  = 2

 

18 + 1/2 + k  = 2

 

(37/2)  + k  = 4/2

 

k =  4/2  - 37/2

 

k = - 33/2

 

 

cool cool cool

 Aug 16, 2019
 #2
avatar+106536 
+1

2) What is the greatest integer value of  b such that -4 is not in the range of  y = x^2 + bx + 12?

 

This parabola turns upward.....therefore...we want to  have the  vertex lie above  ( a, -4)  where a  is the x coordinate of the  vertex....so

 

a = -b / [ 2 (1) ]

2a  =-b

-2a  =b

 

So.....we want to solve this to find the y coordinate of the vertex

 

(a)^2  + (-2a)(a)  + 12  = -4

 

a^2 - 2a^2   = -16

 

-a^2   = -16

 

a^2 =16

 

a = ± 4

 

b achieves its greatest value when a  = -4   →    -2(-4)  = b  = 8

 

Note that  the graph here : https://www.desmos.com/calculator/6vgqqov8mg

shows that   when b  = 8, the  vertex is at  (-4, -4) and when b = 9 that part of the  graph lies below y = -4

But  when b = 7, the graph lies wholly above  y = -4

 

So....b =7  is the greatst integer value that guarantees that -4  is not in the range of  x^2 + bx + 12

 

 

cool cool cool

 Aug 16, 2019
 #3
avatar+106536 
+1

4)  

 

f(x)  = 3x + 3

g(x)  = 4x + 3

 

find  f (g(f(2)))

 

Work from the "inside out"

 

f(2) =   3(2) + 3  =  9

 

g ( f(2))   =  g (9)   =  4(9) + 3  =  39

 

Finally

 

f ( g ( f(2)))  = 

 

f ( 39)  =    3(39) + 3   =  40(3)   = 120

 

 

cool cool cool

 Aug 16, 2019
 #4
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0

THX for the help All of the answers were correct. THX SO MUCH

I still am having troblue with question 3.

 Aug 16, 2019
 #5
avatar+106536 
0

3)  x&  = 7 - x       and       &x  =  x - 7       What is   & (12&)

 

Note that   12&   corresponds to x& = 7 - x....so here, x  must = 12   and we have   7 - 12  =  -5

 

So

 

&(12&)  =  & (-5)

 

And  & (-5)  must correspond to  &x= x - 7  ...so here, x  must =  -5    and we  have  -5 - 7   =  -12

 

So   

 

& (12&) =  - 12

 

 

cool cool cool

 Aug 16, 2019
 #6
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0

THX srry for the late responce. THANK YOU!!laugh

 Aug 16, 2019

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