Punctele in care tangenta la graficul acestei functii este paralela cu axa Ox sunt punctele de extrem din care 2 sint puncte de minim 1 este punct de mazim.
Coordonatele x (abscisele) celor 3 puncte le aflam rezolvand ecuatia:
[tex]\displaystyle\\
f'(x)=0\\
\text{Calculam derivata:}\\\\
f(x)=x^4-8x^2+16\\\\
f'(x)=4x^3-16x=4x(x^2-4)=4x(x-2)(x+2)\\\\
\text{Rezolvam ecuatia:}\\\\
f'(x)=0\\\\
4x(x-2)(x+2)=0\\\\
4x=0\Rightarrow x_1=0\\\\
x-2=0\Rightarrow x_2=2\\\\
x+2=0 \Rightarrow x_3=-2\\\\
\text{inlocuim pe x in functie pentru a-l afla pe y.}\\\\
y_1=f(x_1)=f(0)=0^4-8\cdot0^2+16=16\\\\
y_2=f(x_2)=f(2)=2^4-8\cdot 2^2+16=16-32+16=0\\\\
y_3=f(x_3)=f(-2)=(-2)^4-8\cdot(-2)^2+16=16-32+16=0\\\\
\Rightarrow \boxed{A(0;16),~B(2,0),~C(-2,0)}
[/tex]
