Matematica - Soluzione problema 2 scientifico PNI

[math]
f\left(x\right) =x^3 \ln \left(x\right)\\
\text{Dom}f=\left( 0, +\infty\right)\\
f\left(x\right) > 0\Leftrightarrow \ln\left(x\right)>0 \Leftrightarrow x>1\\
\lim_{x \to 0^+} f\left(x\right)=0\\
\lim_{x \to +\infty} f\left(x\right)=+\infty \text{ nessun asintoto }\\
f'\left(x\right)=3x^2 \ln \left(x \right) + x^3 \frac{1}{x}=\\
3x^2 \ln \left(x\right) + x^2=x^2 \left(3 \ln \left(x\right)+x \right)\geq0 \Leftrightarrow 3\ln \left( x \right) +1 \geq0\\
\ln \left( x \right) \geq -\frac{1}{3} \Rightarrow x \geq e^{-\frac{1}{3}}\\
A \left( e^{-\frac{1}{3}, -\frac{1}{3e}}\right) \text{ minimo assoluto }\\

- f \text{ cresce su }\left(e^{-\frac{1}{3}}, +\infty \right)\\
- f \text{ decresce su } \left(0, e^{-\frac{1}{3}} \right)\\

f''\left(x\right)=2x\left(3\ln\left(x\right) + 1 \right)+ x^2 \frac{3}{x}=\\
6x\ln \left( x \right) + 2x+3x=x\left(6 \ln \left( x \right)+5\right)\geq 0\\
6 \ln \left( x \right)+5 \geq 0\\
x\geq e^{-\frac{5}{6}}
[/math]

[math]
-\frac{1}{3e}\simeq-0.123\\
e^{-\frac{1}{3}}\simeq0.717
[/math]

[math]
P\left(1,0\right)\\
p:y=ax^2+bx+c\\
O \in p \Rightarrow c=0 \Rightarrow y=ax^2+bx\\
\text{inoltre } P\in p \Rightarrow 0=a+b \Rightarrow b=-a\\
\text{Poiché } f'\left(1\right) =1 \text{ ne segue che la tangente alla parabola in p ha equazione } y=x-1\\
\text{Pertanto }

\left\{\begin{array}{l}
y=ax^2-ax\\ y=x-1
\end{array}\right.

\ \Rightarrow\ \ ax^2-(a+1)x+1=0

\text{e quindi}

\Delta=(a+1)^2-4a=(a-1)^2=0\ \Rightarrow\ a=1

\text{da cui l'equazione}

y=x^2-x
[/math]

[math] R=\int_0^1|x^3\cdot \ln x|\ dx=\lim_{a\to 0^+}\int_a -x^3\cdot ln x\ dx=\\
\lim_{a\to 0^+}\left\{\left[\frac{x^4}{4}\ln x\right]_a^1+\int_a^1 \frac{x^4}{4}\cdot \frac{1}{x}\ dx\right\}=\\
\lim_{a\to 0^+}\left\{\frac{a^4}{4}\ln a+\int_a^1\frac{x^3}{4}\ dx\right\}=\\
\lim_{a\to 0^+}\left\{\frac{a^4}{4}\ln a+\left[\frac{x^4}{16}\right]_a^1\right\}=\\
\lim_{a\to 0^+}\left\{\frac{a^4}{4}\ln a+\frac{1}{16}-\frac{a^4}{16}\right\}=\frac{1}{16}[/math]

[math]x=0[/math]

[math]\left\{\begin{array}{l}
x'=-x\\ y'=y
\end{array}\right.[/math]

[math]f_1(x)=-x^3\cdot\ln(-x)[/math]

[math]y=-1[/math]

[math]\left\{\begin{array}{l}
x'=x\\ y'=-y-2
\end{array}\right.[/math]

[math]f_2(x)=-x^3\cdot\ln(x)-2[/math]

[math]f,\ f_1,\ f-2[/math]