Here f(x)=ln(x2+1)−x
We need to find location of all critical points & their nature for the function. For that we need to put f′(x)=0
Here f′(x)=dxd(ln(x2+1)−x)
=x2+12x −1 ∵dxd(ln(x))=x1 & dxd(xn)=nxn−1
To find critical points we will put f′(x)=0
So x2+12x−1=0
⟹x2+12x=1
⟹2x=x2+1
⟹x2−2x+1=0
⟹(x−1)2=0
⟹x=1
So at x=1 we get critical point & when x=1 , f(1)=ln(1+1)−1=ln2−1
So critical point is (1,ln2−1)
Now f′′(x)=dxd(x2+12x−1)=(x2+1)22−2x2
∵dxd(vu)=v2(v.u′−u.v′) & dxd(constant)=0
Now f′′(1)=(2)2(2−2)=40=0
As f′′(1)=0 So we will get inflection point at x=1. That is there is no local Maxima and local minima . (Ans)
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