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
+128408 (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
+128408 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
+128408 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
THX for the help All of the answers were correct. THX SO MUCH
I still am having troblue with question 3.
