Answer to Question #204599 in Calculus for ali

Question #204599

Determine the location of all critical points and determine their nature for the function. f(x)=ln(x^2 + 1) - x


1
Expert's answer
2021-06-10T06:55:57-0400

Here f(x)=ln(x2+1)xf(x)=ln(x^2+1)-x

We need to find location of all critical points & their nature for the function. For that we need to put f(x)=0f'(x)=0


Here f(x)=ddx(ln(x2+1)x)f'(x)=\frac {d} {dx} (ln(x^2+1)-x)


=2xx2+1=\frac {2x} {x^2+1} 1-1 ddx(ln(x))=1x\because \frac {d} {dx} (ln (x))=\frac 1 x & ddx(xn)=nxn1\frac {d} {dx} (x^n)=nx^{n-1}


To find critical points we will put f(x)=0f'(x)=0

So 2xx2+11=0\frac {2x} {x^2+1} -1=0


    2xx2+1=1\implies \frac {2x} {x^2+1} =1

    2x=x2+1\implies 2x=x^2+1

    x22x+1=0\implies x^2-2x+1=0

    (x1)2=0\implies (x-1)^2=0

    x=1\implies x=1

So at x=1x=1 we get critical point & when x=1x=1 , f(1)=ln(1+1)1=ln21f(1)=ln(1+1)-1=ln2-1


So critical point is (1,ln21)(1,ln2 -1)


Now f(x)=ddx(2xx2+11)=22x2(x2+1)2f''(x)=\frac {d} {dx} (\frac {2x} {x^2+1} -1)=\frac {2-2x^2} {(x^2+1)^2}


ddx(uv)=(v.uu.v)v2\because \frac {d} {dx} (\frac u v)=\frac {(v.u'-u.v')} {v^2} & ddx(constant)=0\frac {d} {dx} (constant)=0


Now f(1)=(22)(2)2=04=0f''(1)=\frac {(2-2)} {(2)^2} =\frac 0 4 =0

As f(1)=0f''(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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