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3 years ago

Please solve. Find the Inverse of x Squared f of x equals 6 x + 2

Mathematics

1 answer:

frez [133]3 years ago

8 0

Answer:

The function f of x equals 6 x + 2. First, replace the function notation f of x with y. y equals 6 x plus 2. Reverse the x and y variables. x equals 6 y plus 2. Solve for y by isolating it on one side of the equal sign. First subtract 2 from both sides of the equation. x minus 2 equals 6 y. Divide both sides by 6. The quantity of x minus 2 divided by 6 equals 6 y over 6. The quantity x minus 2 over 6 equals y. Replace y with the inverse function notation. The quantity of x minus 2 divided by 6 equals the inverse function of x. Now solve for the inverse function when x equals two. Substitute 2 for each x. The quantity of 2 minus 2 divided by 6 equals the inverse function of 2. Simplify. 0 divided by 6 equals the inverse function of 2. When x equals two, the corresponding y value on the inverse function is zero. 0 equals the inverse function of 2.

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How do you graph y equals 3 over 2x -4

lesya692 [45]

Answer:

Step-by-step explanation:

3/2 is the slope in y=mx+b format

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3 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Nimfa-mama [501]

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Step-by-step explanation:

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3 years ago

The question is on the image.

irakobra [83]

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