1f(x) = x+2(x+2)2-8(x+2)
= 1/x-6
range of the function = (-∞,+∞) - {0}
8 Answers
1
1venkatesha you have done mistake .
see carefully.
we can approach by y=f(x)
and then for quadratic obtained put D≥0
106y = x+2x2-8x-4 ... domain of x = R- {}
=> yx2 + x(-1-8y) - 4y-2 = 0
as x and y are real,
solving for x should give real solution,
=> x = -(-1-8y) ± √(1+8y)2 + 4(4y+2)2y
now x= 0 gives y = -0.5
using this determine the sign of ± (whether + or - )
then range of f(x) = domain of y in f-1(x) ..
So, u can calculate range of f(x) .. also exclude those values of y which you obtain when denom. of original exp. is zero .
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23y r u making it so complex ?
y = x+2x2 - 8x - 4
i.e x2y -x(8y+1) - (4y+2) = 0
for real x ,
Δ≥0
→ (8y+1)2 + 4 (y)(4y+2) ≥ 0
64y2 +16y + 1 + 16y2+8y ≥ 0
i.e 80y2+24y + 1 ≥ 0
i.e (20y+1)(4y+1) ≥ 0
hence y ε ( - ∞ , -1/4] U [-1/20 ∞)
1708yes querty my solution is more Complex.actually I have done this question in another forum.
I had tried but did,t get (may be some calculation error)
anyway Thanks for urs solution.