domain mania

Domain is all reals.

Now we have to find the increasing-decreasing sections of the domain.

f(x) = 2^x + 2^y

f'(x) = 2^xlog(2) + 2^ylog2 f'(x)
=> f'(x) - 2^ylog2 f'(x) = 2^xlog(2)
=> f'(x) = 2^xlog(2)1 - 2^ylog(2)
=> f'(x) = 2^xlog(2)1 - (2 - 2^x)log(2)

The numerator is always positive. So let's check the denominator.

D_r = 1 - 2log(2) - 2^xlog(2)
This is evidently negative. As f'(x) can never be zero for a real value, it is always negative.
Hence the function is a decreasing one.
As x→(-∞, ∞)
f(x) →(f(Lim_{x→(∞)}x), f(Lim_{x→(-∞)}x))

Now y - 2^y = 2^x
When x→∞, y→∞.
When x→(-∞), y = 2^y => y = log₂ylog(2) = log₂y

So the range is (∞, log₂y) or (log₂y, ∞) when written in the correct format.

Disclaimer : Correctness and accuracy of the above article is in no way guaranteed. And you haven't specified whether f(x) is a different function of x or y itself..

2^y=2 - 2^x

take log on both sides
u get:
y=log(2 - 2^x)log 2

now 2 - 2^x>0
so
x<1 ans

no guts to find out Pritish's mistake..:P

DOMAIN:-

f(x) = 2^x + 2^y= 2

So, 2^y = 2 - 2^x

Therefore, y = log ₁₀ ( 2 - 2^x)

Since, 2 - 2^x > 0
therefore, we get , 2 > 2^x

so, 1 > x or x < 1

so, domain of the function is x belonging to ( - ∞ ,1)

how 'll its graph look like??????????????

erkk......................................................................................
open -∞,1 open