Question

use a computer system or graphing calculator to plot a slope field and/or enough solution curves to indicate whether each critical point of the given differential equation is stable, unstable, or semistable.
$\frac{dx}{dt}=(x^{2}-16)^{2}$
use a computer system or graphing calculator to plot a slope field and solution curves of the equation. choose the correct graph below.
a. image of graph a
b. image of graph b
c. image of graph c
d. image of graph d

Explanation:

Step1: Find critical points

Set $\frac{dx}{dt}=(x^{2}-16)^{2}=0$. Then $x^{2}-16 = 0$, so $x=\pm4$.

Step2: Analyze sign of $\frac{dx}{dt}$

Let $y=(x^{2}-16)^{2}=x^{4}-32x^{2}+256$. The derivative $y' = 4x^{3}-64x=4x(x^{2} - 16)=4x(x - 4)(x + 4)$. For $x<-4$, $\frac{dx}{dt}>0$; for $-40$; for $x>4$, $\frac{dx}{dt}>0$. Since solutions move away from $x = - 4$ and $x = 4$ on one - side and stay away on the other side, the critical points $x=\pm4$ are semistable.

Step3: Match with graph

The graph that shows horizontal tangents at $x = - 4$ and $x = 4$ and the appropriate behavior of the slope field and solution curves is the one where the solution curves do not cross the lines $x=-4$ and $x = 4$ in a way that would indicate stability or instability. By observing the slope - field and solution - curve behavior, we can see that the correct graph is C.

Answer:

C. Option Text