[[Christmas]] New Year Streak 87/88: Calculus Rush #5

Let $f(x)=ax^2+bx+c,$ then $f'(x)=2ax+b$ and $f''(x)=2a.$

Now take a look at $h(x)=f'(x)e^{f(x)}.$ Note that $h'(x)=\{f'(x)^2+f''(x)\}e^{f(x)}.$

There is exactly one root of $h(x)=0,$ call it $x=\alpha.$ We know that $f'(\alpha)=0,$ and thus $h'(\alpha)=f''(\alpha)e^{f(\alpha)}\neq 0.$

Therefore $g(x)=|f'(x)|e^{f(x)}=|h(x)|$ is not differentiable at $x=\alpha.$

Considering $\lim_{x\to\pm\infty} f(x)$ diverges to either infinity, $\lim_{x\to\pm\infty}g(x)$ diverges to either positive infinity or 0.

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From [2.] $g(x)$ has a maximum, which forces $\lim_{x\to\pm\infty}g(x)=0.$ This states that $a<0.$

Then, $h'(x)=\{(2ax+b)^2+2a\}e^{f(x)}$ has two $x$ -intercepts.

These informations let us to realize that $h(x)$ has one local minimum and one local maximum, respectively negative and positive.

This forces $g(x)=|h(x)|$ to only have maxima within where it is differentiable.

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However [3.] states that $g(x)$ has a minimum at $x=2$ . This forces that $\alpha=2,$ leading to $b=-4a.$

Now let $f(x)=ax^2-4ax+c=a(x-2)^2-4a+c.$ We can also say that $f'(x)=2a(x-2).$

For $g(p)=4\sqrt{e}$ we have $g'(p)=h'(p)=0.$

$g(p)=|f'(p)|e^{f(p)}=4\sqrt{e},$ however $p$ is rational, which forces $f'(p)=\pm 4$ and $f(p)=\dfrac{1}{2}.$

$h'(p)=\{f'(p)^2+f''(p)\}e^{f(p)}=0~\Rightarrow~f''(p)=-f'(p)^2~\Rightarrow~2a=-16~\Rightarrow~a=-8.$

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$f(x)=-8(x-2)^2+k$ and $f'(x)=-16(x-2)$ where $f'(p)=-16(p-2)=\pm4$ so that $(p-2)^2=\dfrac{1}{16}.$

$f(p)=-8(p-2)^2+k=\dfrac{1}{2},$ and so $k=1.$

$\therefore~f(x)=-8(x-2)^2+1~\Rightarrow~|f(-1)|=\boxed{71}.$