The function f is such that f(x) = a^2x^2 - ax + 3b for x<=(1/2a), where a and b are constants.
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).
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.
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
Awesome, thank you sooo much Steve and Damon! You guys saved my life!!!
the first guy is correct but the inverse equation should be -1/2 +..... thats all i got the answer sheet so
could you elaborate on your answers. how did you get them?
Your answer is wrong.
The answer is -13/4
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