How do I find the range of the function #y=-2^x+2#?

Question

Madelyn Anderson · Accepted Answer

Here's my attempt: #y<2# is the range. #y=-2^x+2# Can also be written as #2-y=2^x# and also, #ln(2-y)=xln2# #=> 2-y>0 => y<2#

Elijah Allen · Answer

The range of #y=-2^x+2# is #(-infty;2)#. I'd start from a known fact that #a^x>0# for all #a>0# and #x in RR#. So: #2^x>0# #-2^x<0# #-2^x+2<2# #y<2# #y in (-infty;2)#

Madelyn Anderson · Answer

undefined To find the range of the function $ y = -2^x + 2 $, we first need to understand the behavior of the exponential function $ -2^x $.

The range of an exponential function $ a^x $, where $ a $ is a constant and $ x $ varies over all real numbers, depends on the value of $ a $:

- If $ 0 < a < 1 $, the range is $ (0, +\infty) $, meaning the function's values are positive and approach zero but never reach it.
- If $ a = 1 $, the range is $ \{1\} $, because $ 1^x = 1 $ for all real numbers $ x $.
- If $ a > 1 $, the range is $ (-\infty, +\infty) $, meaning the function's values cover all real numbers.

In this case, $ a = -2 $, so $ -2^x $ will have all negative values except when $ x $ is even, in which case it will be positive.

Adding 2 to $ -2^x $ shifts the graph vertically upward by 2 units.

Thus, the range of the function $ y = -2^x + 2 $ is all real numbers, except that it's shifted upward by 2 units, so it does not reach values less than 2. Therefore, the range is $ (-\infty, 2] $.