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3 years ago

Describe the points with a double inequality. I need the answer quick. Thanks!

Mathematics

1 answer:

Dominik [7]3 years ago

8 0

Answer:

$- 3 \: < \: x \: < \: 13$

Step-by-step explanation:

We want to describe the solution set on the number line as an inequality.

We have open circles at both ends.

This means the inequalities are not the ones containing equal sign.

The number line started from -3 and ended at 13 non inclusive.

This means the solution set for the inequality are all real numbers greater than -3 but less than 13.

We write this as:

$- 3 \: < \: x \: < \: 13$

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ddd [48]

Check where the first-order partial derivatives vanish to find any critical points within the given region:

$f(x,y)=2x^3+y^4\implies\begin{cases}f_x=6x^2=0\\f_y=4y^3=0\end{cases}\implies(x,y)=(0,0)$

The Hessian for this function is

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}12x&0\\0&12y^2\end{bmatrix}$

with $\det\mathbf H(0,0)=0$ , so unfortunately the second partial derivative test fails. However, if we take $x=0$ we see that $f(x,y)>0$ for different values of $y$ ; if we take $y=0$ we see $f(x,y)$ takes on both positive and negative values. This indicates (0, 0) is neither the site of an extremum nor a saddle point.

Now check for points along the boundary. We can parameterize the boundary by

$(x,y)=(8\cos t,8\sin t)$

with $0\le t\le2\pi$ . This turns $f(x,y)$ into a univariate function $F(t)$ :

$F(t)=f(8\cos t,8\sin t)=2^{10}\cos^3t+2^{12}\sin^4t$

$\implies F'(t)=3\cdot2^{10}\cos^2t(-\sin t)+2^{14}\sin^3t\cos t=2^{10}\sin t\cos t(16\sin^2t-3\cos t)$

$F'(t)=0\implies\begin{matrix}\sin t=0\implies t=0,\,t=\pi\\\\\cos t=0\implies t=\dfrac\pi2,\,t=\dfrac{3\pi}2\\\\16\sin^2t-3\cos t=0\implies t=2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3},\,t=2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\end{matrix}$

At these critical points, we get

$F(0)=1024$

$F\left(2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893$

$F\left(\dfrac\pi2\right)=4096$

$F(\pi)=-1024$

$F\left(\dfrac{3\pi}2\right)=4096$

$F\left(2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893$

We only care about 3 of these results.

$t=\dfrac\pi2\implies(x,y)=(0,8)$

$t=\pi\implies(x,y)=(-8,0)$

$t=\dfrac{3\pi}2\implies(x,y)=(0,-8)$

So to recap, we found that $f(x,y)$ attains

a maximum value of 4096 at the points (0, 8) and (0, -8), and
a minimum value of -1024 at the point (-8, 0).

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