Which equations are true?A: -x = -x + x + 0B: -x = 0 + (-x)C: None Of The Above

Mathematics

1 answer:

Leto [7]2 years ago

3 0

Answer: B
Explanation: if you look at A and try to combine like terms, you will add the -x on the right side the the one on the left, cancelling out the x on the left. then you have 0=x+0, and unless x=0, it is not true.
With B, when you add x to the -x, you cancel it out, giving you 0=0, which is true. hope it helps!

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Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$