Question

The range of the function f(x) = e^x + e^-x + (pi) ^ x + (pi) ^-x

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Answer 1

Answer:

Answer:

decreasing on [0,∞)

Step-by-step explanation:

f(x)=log(e+x)log(π+x)

∴f′(x)=(log(e+x))2(π+x1)log(e+x)−(e+x1)log(π+x)

Let h(x)=π+xlog(e+x)−e+xlog(π+x)

Let us consider g(x)=xlnx

⇒g′(x)=x×x1+lnx=1+lnx

g′(x)>0∀x∈(e1,∞) and g′(x)<0∀x∈(0,e1)

Now e<π⇒e+x<π+

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