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If f\left ( x \right )=log\left ( x^{2}+1 \right ) and g\left ( x \right )=e^{x}, then fog\left ( x \right ) is

Option: 1

x^{2}+1

 

 

 


Option: 2

log\left ( e^{2x}+1 \right )


Option: 3

e^{2x}+1


Option: 4

2x+1


Answers (1)

best_answer

f \circ g(x)=f(g(x))=\log \left((g(x))^{2}+1\right)=\log \left(e^{2 x}+1\right)

Posted by

seema garhwal

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