Exercise 8.18

Question

Exercise 18: 
    Define
    \begin{align*}
        f(x) &= x^3-\sin^2x	an x \
        g(x) &= 2x^2-\sin^2x-x	an x.
    \end{align*}
    Find out, for each of these two functions, whether it is positive or negative for all $x\in(0,\pi/2)$, or whether it changes sign. Prove your answer.

analambanomenos · Accepted Answer

Note that $(d/dx)	an x=n	an^{n-1}x+n	an^{n+1}x$.
    \begin{align*}
        f(x) &= x^3-(1-\cos^2x)	an x \
        &= x^3+2^{-1}\sin2x-	an x \
        f'(x) &= 3x^2 +\cos2x-1-	an^2x \
        f''(x) &= 6x-2\sin2x-2	an x-2	an^3x \
        f'''(x) &= 6-4\cos2x-2-8	an^2x-6	an^4x \
	f^{(4)}(x) &= 8\sin2x-16	an x-40	an^3x-24	an^5x \
	f^{(5)}(x) &= 16\cos2x-16-136	an^2x-240	an^4x-120	an^6x \
	f^{(6)}(x) &= -32\sin2x-272	an x-1232	an^3x-1680	an^5x-720	an^7x
    \end{align*}
    Note that $f(0)=f'(0)=f''(0)=f'''(0)=f^{(4)}(0)=f^{(5)}(0)=0$. Since $f^{(6)}(x)<0$ on $(0,\pi/2)$, $f^{(5)}$ decreases monotonically throughout that interval from $f^{(5)}(0)=0$ to 
    $f^{(5)}(\pi/2)=-\infty$, hence the same is successively true for $f^{(4)}$, $f'''$, $f''$, $f'$, and $f$.
    \begin{align*}
        g(x) &= 2x^2-\sin^2x-x	an x \
        g'(x) &= 4x-\sin2x-	an x-x(1+	an^2x) \
        g''(x) &= 4-2\cos2x-2-2	an^2x-x(2	an x+2	an^3x) \
        g'''(x) &= 4\sin2x-6	an x-6	an^3x-x(2+8	an^2x+6	an^3x) \
        g^{(4)}(x) &= 8\cos2x-8-32	an^2x-24	an^4x-x(16	an x+40	an^3x+24	an^5x) \
        g^{(5)}(x) &= -16\sin2x-80	an x-200	an^3x-120	an^5x\;-\
        &\phantom{=}\;\; x(16+136	an^2x+240	an^3x+120	an^6x)
    \end{align*}
    Note that $g(0)=g'(0)=g''(0)=g'''(0)=g^{(4)}(0)=0$. Since $g^{(5)}(x)<0$ on $(0,\pi/2)$, $g^{(4)}$ decreases monotonically throughout that interval from $g^{(4)}(0)=0$ to $g^{(4)}(\pi/2)=-\infty$,
    hence the same is successively true for $g'''$, $g''$, $g'$, and $g$.

Exercise 8.18

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Exercise 8.18

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