Exercise 5.15

Let $g(x)=A∕x+\mathit{Bx}$ for $x>0$ where $A$ and $B$ are positive real numbers. Then $g^{\prime }(x)=-A∕x^2+B$ and $g^{″}(x)=2A∕x^3>0$. Since $g^{\prime }(x)=0$ for $x=\sqrt{A∕B}$, $g$ has the minimum value $g\left(\sqrt{A∕B}\right)=2\sqrt{\mathit{AB}}$.

Following the hint, by Theorem 5.15, for $h>0$ there is $\xi \in (x,x+2h)$ such that

Hence $|f^{\prime }(x)|\le M_0∕h+M_{2}h\le 2\sqrt{M_0{M}_2}$ by the previous result, or $M_1^2\le 4M_0{M}_2$.

Letting $f(x)$ be the example above, we get

For $x<0$, $f^{\prime }$ is negative, so $f(x)$ decreases from 1 to $-1$, and for $x>0$, $f^{\prime }$ is positive, so $f(x)$ increases monotonically from $-1$ to a limit of $1$. Hence $M_0=1$.

For $x<0$, $f^{\prime }$ increases linearly from $-4$ to 0. For $x>0$, $f^{″}(x)=0$ has a single solution at $x=\sqrt{3}∕3$, so $f^{\prime }(x)$ has a maximum value of $3\sqrt{3}∕4$. Hence $M_1=4$.

For $x>0$, $f^{(3)}(x)$ has the single solution $x=1$, so $f^{″}(x)$ decreases from 4 to a minimum value of $f^{″}(1)=-1$, then increases monotonically to a limit of 0. Hence $M_2=4$.

Hence for this example, $M_1^2=4M_0{M}_1=16$.