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

Guest Aug 16, 2019

#1**+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

CPhill Aug 16, 2019

#2**+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

CPhill Aug 16, 2019

#3**+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

CPhill Aug 16, 2019

#4**0 **

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

I still am having troblue with question 3.

Guest Aug 16, 2019