Range of a function.

venkatesha you have done mistake .
see carefully.

we can approach by y=f(x)

and then for quadratic obtained put D≥0

wht mistake..... plz tell me........

y = 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 .

y 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 ∞)

yes 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.