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Question: How do I find the range of the function \(y = - {2^x} + 2\)?...

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

Explanation

Solution

We have to determine the range of the given function. For this, put y=f(x)y = f\left( x \right) and solve the equation y=f(x)y = f\left( x \right) for xx in terms of yy. Next, put x=ϕ(y)x = \phi \left( y \right) and find the values of yy for which the values of xx, obtained from x=ϕ(y)x = \phi \left( y \right), are real and in the domain of ff. Thus, the set of values of yy obtained is the range of ff.

Complete step by step solution:
Given function: y=2x+2y = - {2^x} + 2
We have to find the range of a given function.
For this, put y=f(x)y = f\left( x \right) and solve the equation y=f(x)y = f\left( x \right) for xx in terms of yy.
y=2x+2\Rightarrow y = - {2^x} + 2
It can be written as 2x=2y{2^x} = 2 - y
Take logarithm both sides of the equation, we get
ln(2x)=ln(2y)\Rightarrow \ln \left( {{2^x}} \right) = \ln \left( {2 - y} \right)
Use property ln(am)=mln(a)\ln \left( {{a^m}} \right) = m\ln \left( a \right), we get
xln(2)=ln(2y)\Rightarrow x\ln \left( 2 \right) = \ln \left( {2 - y} \right)
Now, put x=ϕ(y)x = \phi \left( y \right) and find the values of yy for which the values of xx, obtained from x=ϕ(y)x = \phi \left( y \right), are real and in the domain of ff.
2y>02 - y > 0
y<2\Rightarrow y < 2
Clearly, Range (ff) =(,2) = \left( { - \infty ,2} \right).

Therefore, the range of given function is (,2)\left( { - \infty ,2} \right).

Note: In above question, we can determine the range of a given question by simply drawing the graph of the function.

From the graph, we can observe that y<2y < 2 for all values of xx.
Therefore, the range of the given function is (,2)\left( { - \infty ,2} \right).