34 multiple choice 1 point the slope field for a certain differential equation is shown above. which of the…

Answer:

E. $y = \frac{1}{1 + x^{2}}$

Explanation:

Step1: Check initial - condition for each option

For option A: When $x = 0$, $y=\cos(0)=1$. For option B: When $x = 0$, $y=1-0^{2}=1$. For option C: When $x = 0$, $y = e^{0}=1$. For option D: When $x = 0$, $y=\sqrt{1 - 0^{2}}=1$. For option E: When $x = 0$, $y=\frac{1}{1+0^{2}}=1$. All options satisfy $y(0)=1$.

Step2: Analyze the behavior of the slope - field

The slope - field shows that the function is symmetric about the y - axis and is decreasing for $x>0$ and increasing for $x < 0$. For option A: $y=\cos x$, $y'=-\sin x$. At $x = 0$, $y' = 0$, but for small positive $x$, $y'$ is negative and it oscillates, not consistent with the non - oscillatory slope field. For option B: $y = 1-x^{2}$, $y'=-2x$. It is a parabola - shaped derivative, and the rate of change is linear, not consistent with the slope field's curvature. For option C: $y = e^{x}$, $y'=e^{x}$. It is always increasing for all $x$, which is not consistent with the slope field that shows a decreasing behavior for $x>0$. For option D: $y=\sqrt{1 - x^{2}}$, $y'=\frac{-x}{\sqrt{1 - x^{2}}}$, its domain is $- 1\leqslant x\leqslant1$. For option E: $y=\frac{1}{1 + x^{2}}$, $y'=\frac{-2x}{(1 + x^{2})^{2}}$. It is symmetric about the y - axis, increasing for $x<0$ and decreasing for $x > 0$, which is consistent with the slope field.