The function g(x)= x^ square root of 1+x is neither an even function nor an odd function
How to determine the function type?
The function is given as
g(x)= x^ square root of 1+x
Rewrite the above function equation properly
So, we have
g(x) = x^(√x+1)
Calculate g(-x)
So, we have
g(-x) = -x^(√-x+1)
Evaluate
g(-x) = -x^(√-x+1)
Calculate -g(x)
So, we have
-g(x) = -x^(√x+1)
In the above computations;
g(x) does not equal g(-x) and -g(x)
Hence, the function g(x)= x^ square root of 1+x is neither an even function nor an odd function
Read more about function types at:
brainly.com/question/23934926
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