Graph the polynomial function.

$f\left(x\right)={x}^{5}+{x}^{2}-4$

kuCAu
2021-08-06
Answered

Graph the polynomial function.

$f\left(x\right)={x}^{5}+{x}^{2}-4$

You can still ask an expert for help

Isma Jimenez

Answered 2021-08-07
Author has **84** answers

Step 1

$f\left(x\right)={x}^{5}+{x}^{2}-4$

The leading coefficient is 1 (positive) and the degree is 5 (odd)

Therefore$f\left(x\right)\to \mathrm{\infty}$ , as $x\to \mathrm{\infty}$

and$f\left(x\right)\to -\mathrm{\infty}$ , as $x\to -\mathrm{\infty}$

Step 2

The leading coefficient is 1 (positive) and the degree is 5 (odd)

Therefore

and

Step 2

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What is the Galois group of $27{x}^{8}-72{x}^{4}-16$ over the rationals?

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The roots of ${x}^{3}+4x-1=0$ are a,b,c.

Find$(a+1)}^{-3}+{(b+1)}^{-3}+{(c+1)}^{-3$

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asked 2022-07-15

Calculate trigonometry expression: $\frac{\mathrm{cos}3\alpha -\mathrm{sin}3\alpha}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}$ if $\frac{\mathrm{cos}3\alpha -\mathrm{sin}3\alpha}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\textstyle \text{if}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}({\displaystyle \frac{\pi}{4}}-\alpha )=0,1.$

I make following transformation:

$\begin{array}{c}{\displaystyle \frac{\mathrm{cos}3\alpha -\mathrm{sin}3\alpha}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}={\displaystyle \frac{4{\mathrm{cos}}^{3}\alpha -3\mathrm{cos}\alpha -3\mathrm{sin}\alpha +4{\mathrm{sin}}^{3}\alpha}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=\\ ={\displaystyle \frac{4({\mathrm{cos}}^{3}\alpha +{\mathrm{sin}}^{3}\alpha )-3(\mathrm{cos}\alpha +\mathrm{sin}\alpha )}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=\\ ={\displaystyle \frac{4(\mathrm{cos}\alpha +\mathrm{sin}\alpha )({\mathrm{cos}}^{2}\alpha -\mathrm{cos}\alpha \cdot \mathrm{sin}\alpha +{\mathrm{sin}}^{2}\alpha )-3(\mathrm{cos}\alpha +\mathrm{sin}\alpha )}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=\\ {\displaystyle \frac{(\mathrm{cos}\alpha +\mathrm{sin}\alpha )[\phantom{\rule{thickmathspace}{0ex}}4\phantom{\rule{thickmathspace}{0ex}}(1-\mathrm{sin}\alpha \mathrm{cos}\alpha )-3\phantom{\rule{thickmathspace}{0ex}}]}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=4-4\mathrm{sin}\alpha \mathrm{cos}\alpha -3=1-2\mathrm{sin}2\alpha .\end{array}$

What next? How i can use substitution?

I make following transformation:

$\begin{array}{c}{\displaystyle \frac{\mathrm{cos}3\alpha -\mathrm{sin}3\alpha}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}={\displaystyle \frac{4{\mathrm{cos}}^{3}\alpha -3\mathrm{cos}\alpha -3\mathrm{sin}\alpha +4{\mathrm{sin}}^{3}\alpha}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=\\ ={\displaystyle \frac{4({\mathrm{cos}}^{3}\alpha +{\mathrm{sin}}^{3}\alpha )-3(\mathrm{cos}\alpha +\mathrm{sin}\alpha )}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=\\ ={\displaystyle \frac{4(\mathrm{cos}\alpha +\mathrm{sin}\alpha )({\mathrm{cos}}^{2}\alpha -\mathrm{cos}\alpha \cdot \mathrm{sin}\alpha +{\mathrm{sin}}^{2}\alpha )-3(\mathrm{cos}\alpha +\mathrm{sin}\alpha )}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=\\ {\displaystyle \frac{(\mathrm{cos}\alpha +\mathrm{sin}\alpha )[\phantom{\rule{thickmathspace}{0ex}}4\phantom{\rule{thickmathspace}{0ex}}(1-\mathrm{sin}\alpha \mathrm{cos}\alpha )-3\phantom{\rule{thickmathspace}{0ex}}]}{\mathrm{cos}\alpha +\mathrm{sin}\alpha}}=4-4\mathrm{sin}\alpha \mathrm{cos}\alpha -3=1-2\mathrm{sin}2\alpha .\end{array}$

What next? How i can use substitution?

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$\int \mathrm{cos}\left(2{\mathrm{cot}}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$

I proceeded as follows:

->first, let$x=\mathrm{cos}\left(2\theta \right)\Rightarrow dx=-2\mathrm{sin}2\theta d\theta$ so the integral becomes:

$\int \mathrm{cos}\left(2{\mathrm{cot}}^{-1}\sqrt{\frac{1-\mathrm{cos}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}}\right)-2\mathrm{sin}2\theta d\theta =\int \mathrm{cos}\left(2{\mathrm{cot}}^{-1}\sqrt{\frac{{\mathrm{sin}}^{2}\theta}{{\mathrm{cos}}^{2}\theta}}\right)-2\mathrm{sin}2\theta d\theta$

$=\int \mathrm{cos}(2{\mathrm{cot}}^{-1}\left(\mathrm{tan}\theta \right)-2\mathrm{sin}2\theta d\theta )=\int \mathrm{cos}\frac{2}{\theta}-2\mathrm{sin}2\theta d\theta$

After this i am stuck. How do i proceed?

I proceeded as follows:

->first, let

After this i am stuck. How do i proceed?

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