[[Christmas]] New Year Streak 86/88: Calculus Rush #4

Since $g(x)=x+\ln f(x),$ we have $g'(x)=1+\dfrac{f'(x)}{f(x)}.$

Now let $f(x)=ax^2+bx+1,$ then because of [1.] we have $b^2<4a$ and $a>0.$

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From [2.] we see that $1+\dfrac{f'(x)}{f(x)}\ge 0,$ which leads us to $f(x)+f'(x)\ge0.$

$ax^2+(2a+b)x+b+1\ge0~\Leftrightarrow~D=4a^2+b^2-4a\le 0.$

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In [3.] , $\left|g(x)-g'(0)x\right|=\left|x+\ln f(x)-x-\dfrac{f'(0)}{f(0)}x\right|=\left|\ln (ax^2+bx+1)-bx\right|$

Letting $x=0,$ we note that two graphs $y=\ln (ax^2+bx+1)$ and $y=bx$ intersects at the origin.

Then let $h(x)=\ln(ax^2+bx+1).$

$h''(x)=\dfrac{-2a^2x-2abx-b^2+2a}{(ax^2+bx+1)^2},$ and $D_{/4}=(ab)^2-2a^2\cdot (b^2-2a)=a^2(-b^2+4a)>0.$

Therefore $y=h(x)$ has two inflection points.

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$h(x)$ is a logarithmic function, which means as $x$ tends to either infinity, the graph is concave.

This tells us that the abstract shape of $h(x)$ would be $[-\infty~\leftarrow~\text{concave}]-[\text{convex}]-[\text{concave}~\rightarrow~\infty].$

So for $|h(x)-bx|$ to be differentiable over all reals, we need to have $y=bx$ tangent to $y=h(x)$ at the local minimum, or at the inflection point.

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$i)$ tangent at the local minimum

Then $b=0,$ leading to $f(x)=ax^2+1\ge 1,$ telling us that $f(x)\ge 1.$

$ii)$ tangent at the inflection point

Then $y=bx$ is the inflection tangent line of $y=h(x)$ at the origin, and from $h'(0)=b$ (which is pretty useless) and $h''(0)=0,$ we get that $b^2=2a.$

From before we found that $4a^2+b^2-4a\le 0.$ Plug that in, and we get $2a^2-a\le 0~\Leftrightarrow~ 0<a\le \dfrac{1}{2}$ and $0<b^2\le 1.$

Since $f(k)=ak^2+bk+1=\dfrac{1}{2}b^2k^2+bk+1=\dfrac{1}{2}\left(bk+1\right)^2+\dfrac{1}{2},$ for all $k$ the minimum of $f(k)$ is $\dfrac{1}{2}.$ (automatically discarding $i).$ )

$\therefore~\displaystyle \sum_{n=1}^{2017}h(n)^3 = 2017\cdot \dfrac{1}{8}=\boxed{252.125}.$