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3 years ago

Find the inverse of the following function

Mathematics

2 answers:

brilliants [131]3 years ago

8 0

Answer:

ƒ^-1 (x) = $\frac{2-7x}{2+x}$

Step-by-step explanation:

Substitute y for f (x)

$y = \frac{-2x+2}{x+7}$

Interchange x and y

$x=\frac{-2y+2}{y+7}$

Swap the sides of the equation

$\frac{-2y+2}{y+7} = x$

Multiply both sides of the equation by y + 7

-2y + 2 = (y + 7)x

Distribute x through the parentheses

-2y + 2 = xy + 7x

Move the expression/constant to the left-hand side and change its sign

-2y - xy + 7x - 2

Factor out from the expression

(-2 - x)y = 7x - 2

Divide both sides of the equation by -2 - x

$y = \frac{7x-2}{-2-x}$

Simplify the expression

$y = \frac{2-7x}{2+x}$

Substitute ƒ ^-1 (x) for y

$f^{-1} (x) = \frac{2-7x}{2+x}$

KatRina [158]3 years ago

3 0

f-¹(x)= $\frac{2-7x}{x+2}$

Answer:

Solution given:

f(x)= $\frac{-2x+2}{x+7}$

Let f(x)=y

y= $\frac{-2x+2}{x+7}$

Interchanging role of x and y

x= $\frac{-2y+2}{y+7}$

doing crisscrossed multiplication

x(y+7)=-2y+2

now solve it:

xy+7x=-2y+2

keep like terms in one side

xy+2y=2-7x

take common

y(x+2)=2-7x

make a value of y

y= $\frac{2-7x}{x+2}$

So,

f-¹(x)= $\frac{2-7x}{x+2}$

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$\boxed{\sf - 2 {p}^{2} - 11p - 35}$

Step-by-step explanation:

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Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

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We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

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Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

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$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

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$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

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Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

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Find the inverse of the following function​

Find the inverse of the following function