The function f is such that f(x) = a^2x^2 - ax + 3b for x<=(1/2a), where a and b are constants.

Question

1) For the case where f(-2) = 4a^2 - b +8 and f(-3) = 7a^2 - b + 14, find the possible values of a and b.
2) For the case where a = 1 and b = -1, find an expression for inverse f(x) and give the domain of inverse f(x).

Answer 1

1)
You have two equations using f(-2) and f(-3):

4a^2+2a+3b = 4a^2-b+8
9a^2+3a+3b = 7a^2-b+14

Those reduce to

a+2b = 4
2a^2+3a+4b = 14

Solve those and you get
(a,b) = (3/2,5/4) or (-2,3)

(2) a=1 b = -1 means

f(x) = x^2-x+6 = (x - 1/2)^2 + 23/4
The range is y >= 23/4

so, f^-1(x) = 1/2 + √(x - 23/4)
for x <= 1/2
with domain x >= 23/4

Answer 2

do you mean 1/(2a) or (1/2)a

1)
f(-2) = 4 a^2 +2a + 3b = 4 a^2-b+8

f(-3) = 9 a^2 +3a + 3b = 7a^2-b+14

so 2 a + 3 b = - b + 8
2 a + 4 b = 8
a + 2 b = 4 or a = 4 - 2 b
and then
2 a^2 + 3 a + 4 b = 14
2(16 - 16 b +4b^2) + 3(4-2b) +4b=14
32 -32b +8b^2+12 -6b +4b = 14

30 + 8b^2 -34b = 0

4 b^2 - 17 b + 15 = 0
b = 5/4 or b = 3
then a = 3/2 or a = -2

check that x < whatever you wrote for those two points

Answer 3

Awesome, thank you sooo much Steve and Damon! You guys saved my life!!!

Answer 4

could you elaborate on your answers. how did you get them?

Answer 5

Your answer is wrong.
The answer is -13/4

Answer 6

the first guy is correct but the inverse equation should be -1/2 +..... thats all i got the answer sheet so

Answer 7

When a=1 and b=-1.(x-1/2)=(y-13/4)^1/2
or, x=1/2)+(y-13/4)^1/2 But if y=f(x) then x=f^-1(y)
Then f^-1(y)=1/2)+(y-13/4)^1/2.
Replace dummy variable y by x to make variable seen as x as that of question.
then the above expression becomes f^-1(x)=1/2)+(x-13/4)^1/2.
The range will be real valued when domain will be finite which is possible when x-13/4>0 and x-13/4=0
There x=13/4.
Ans domain AS CIE Level Question.