Real Analysis 513fa(x)−f 0 (x)={
0 forx = 0 ,
a forx= 0 ,does not have the intermediate value property, so it is not the derivative of a function,
fa(x)itself cannot be the derivative of a function. This completes the solution.
(Romanian high school textbook)
404.Define the functionf:R→R,f(x)=ex−x−1. Its first derivativef′(x)=ex− 1
has the unique zerox =0, and the second derivativef′′(x)=exis strictly positive.
It follows thatx=0 is a global minimum off, and becausef( 0 )=0,f(x) >0 for
x =0. Hence the inequality.
405.Taking the logarithm, transform the equation into the equivalentxln 2=2lnx.
Define the functionf:R→R,f(x)=xln 2−2lnx. We are to find the zeros off.
Differentiating, we obtain
f′(x)=ln 2−2
x,
which is strictly increasing. The unique zero of the derivative isln 2^2 , and sof′is negative
forx< 2 /ln 2 and positive forx>ln 2^2. Note also that limx→ 0 f(x)=limx→∞f(x)=
∞. There are two possibilities: eitherf(ln 2^2 )>0, in which case the equationf(x)= 0
has no solutions, orf(ln 2^2 )<0, in which case the equationf(x)=0 has exactly two
solutions. The latter must be true, sincef( 2 )=f( 4 )=0. Therefore,x=2 andx= 4
are the only solutions tof(x)=0, and hence also to the original equation.
406.Iff(x)≥ 0 for allx, then the functiong(x)= (x−a 1 )(x−a 2 )(x−a 3 )is
increasing, since its derivative isf. It follows thatghas only one zero, and we conclude
thata 1 =a 2 =a 3.
(V. Boskoff)
407.Letf:C→C,f(z)=z^3 −z+2. We have to determine max|z|= 1 |f(z)|^2. For
this, we switch to real coordinates. If|z|=1, thenz=x+iywithy^2 = 1 −x^2 ,
− 1 ≤x≤1. View the restriction of|f(z)|^2 to the unit circle as a function depending on
the real variablex:
|f(z)|^2 =|(x+iy)^3 −(x+iy)+ 2 |^2
=|(x^3 − 3 xy^2 −x+ 2 )+iy( 3 x^2 −y^2 − 1 )|^2
=|(x^3 − 3 x( 1 −x^2 )−x+ 2 )+iy( 3 x^2 −( 1 −x^2 )− 1 )|^2
=( 4 x^3 − 4 x+ 2 )^2 +( 1 −x^2 )( 4 x^2 − 2 )^2
= 16 x^3 − 4 x^2 − 16 x+ 8.Call this last expressiong(x). Its maximum on[− 1 , 1 ]is either at a critical point or at an
endpoint of the interval. The critical points are the roots ofg′(x)= 48 x^2 − 8 x− 16 =0,