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4 years ago

Consider the function f(x) = x2. Which of the following functions shifts f(x)

Mathematics

1 answer:

tresset_1 [31]4 years ago

6 0

Answer:

f(x) = (x - 3)² - 5

Step-by-step explanation:

equate equation to 0

(x - 3)² = 0

take the square root on both sides

x - 3 = 0

add 3

x = 3

If x = 3 then you are moving to 3 units to the right.

- 5 means you are going downward 5 units.

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Masteriza [31]

First substitute $x=\tan^{-1}(y)$ to rewrite the integral as

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = \int_0^1 \frac{\sqrt{y-y^2}}{1+y^2} \, dy$

Now use an Euler substitution, $z=\frac{\sqrt{y-y^2}}y$ to rewrite it again as

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = 2 \int_0^\infty \frac{t^2}{(t^2+1)^2 + 1) (t^2 + 1)} \, dt$

where we take

$\sqrt{y - y^2} = \sqrt{-y(y-1)} = yt \implies y = \dfrac1{1+t^2} \text{ and } dy = -\dfrac{2t}{(1+t^2)^2} \, dt$

Partial fractions:

$\displaystyle \frac{t^2}{((t^2+1)^2+1) (t^2 + 1)} = \dfrac{t^2+2}{t^4+2t^2+2} - \dfrac1{t^2+1}$

so that

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = 2 \left(\int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt - \int_0^\infty \frac{dt}{t^2+1}\right)$

The second integral is trivial,

$\displaystyle \int_0^\infty \frac{dt}{t^2+1} = \lim_{t\to\infty}\tan^{-1}(t) - \tan^{-1}(0) = \frac\pi2$

For the other, I'm compelled to use the residue theorem, though real methods are doable too (e.g. trig substitution). Consider the contour integral

$\displaystyle \int_\Gamma f(z) \, dz = \int_\Gamma \frac{z^2+2}{z^4+2z^2+2} \, dz$

where $\Gamma$ is a semicircle in the upper half of the complex plane, and its diameter lies on the real axis connecting $-R$ to $R$ . The value of this integral is 2πi times the sum of the residues in the upper half-plane. It's fairly straightforward to convince ourselves that the integral along the circular arc vanishes as $R\to\infty$ , so the contour integral converges to the integral over the entire real line. Note that

$\displaystyle 2 \int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt = \int_{-\infty}^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt$

since the integrand is even.

Find the poles of $f(z)$ .

$z^4 + 2z^2 + 2 = 0 \\\\ ~~~~ \implies (z^2+1)^2 = -1 \\\\ ~~~~ \implies z^2 = -1 \pm i \\\\ ~~~~ \implies z = \pm \sqrt{-1 \pm i} = \sqrt[4]{2}\, e^{\pm i(3\pi/8 + \pi k)}$

where $k\in\{0,1\}$ .

The two poles we care about are at $z_1=\sqrt[4]{2}\,e^{i\,3\pi/8}$ and $z_2=\sqrt[4]{2}\,e^{-i\,11\pi/8}$ . Compute the residues at each one.

$\displaystyle \mathrm{Res}\left\{f(z),z=z_1\right\} = \lim_{z\to z_1} \frac{f(z)}{z-z_1} = -\frac1{2^{7/4}}\,ie^{-i\,\pi/8} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~= -\frac1{2^{7/4}} \left(\sin\left(\frac\pi8\right) + i \cos\left(\frac\pi8\right)\right)$

$\displaystyle \mathrm{Res}\left\{f(z),z=z_2\right\} = \lim_{z\to z_2} \frac{f(z)}{z-z_2} = -\frac1{2^{7/4}}\,ie^{i\,\pi/8} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~= \frac1{2^{7/4}} \left(\sin\left(\frac\pi8\right) - i \cos\left(\frac\pi8\right)\right)$

By the residue theorem,

$\displaystyle \int f(z) \, dz = 2\pi i \sum_{\rm poles} \mathrm{Res}\{f(z)\} = \frac{4\pi}{2^{7/4}} \cos\left(\frac\pi8\right)$

We also have

$\displaystyle \cos^2\left(\dfrac\pi8\right) = \dfrac{1 + \cos\left(\frac\pi4\right)}2 = \dfrac{2 + \sqrt2}4 \implies \cos\left(\frac\pi8\right) = \dfrac{\sqrt{2+\sqrt2}}2$

Then the remaining integral is

$\displaystyle \int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt = \frac{4\pi}{2^{7/4}} \cos\left(\frac\pi8\right) = \sqrt{\frac12 + \frac1{\sqrt2}} \, \pi$

It follows that

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = \boxed{\left(\sqrt{\frac12 + \frac1{\sqrt2}} - 1\right) \pi}$

7 0

1 year ago

Solve the solution using the substitution method <br><br>3x +5y = 29<br>y= -4x -1

Artist 52 [7]

3x + 5y = 29 first equation
y = -4x - 1 second equation

substitute the second equation to the first equation;
3x + 5(-4x - 1) = 29
-17x - 5 = 29

solve for x;
-17x - 5 = 29
-17x = 34
x = -2

substitute x = -2 into the second equation to find y;
y = -4x - 1
y = -4(-2) - 1
y = 8 - 1
y = 7

so;
x = -2, y = 7

hope this helps, God bless!

3 0

3 years ago

50 Points!!! Easy I’m just dumb!!!!

natta225 [31]

Answer:

20.9-26.2 = -5.3 to the left (or 5.3 to the right)

6 0

3 years ago

Read 2 more answers

A number x is at least 5 and is at most 8

Rainbow [258]

Answer:

$5 \leqslant x \leqslant 8$

7 0

2 years ago

Jemina flips a disc of the type shown below: A disc with two layers is shown. The top layer is shaded in a darker shade of gray,

Yakvenalex [24]

The probability is just 1/2 that the disc will result with the white side landing up.

There are two sides to the disc. If they are equally likely to happen, then the chance of either side is 1/2. It does not matter which number the flip is.

6 0

3 years ago