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Question

The inverse of the function $y equals open square brackets 1 minus left parenthesis x minus 3 right parenthesis to the power of 4 end exponent close square brackets to the power of 1 divided by 7 end exponent$ is

$3 + {(1 - x^{7})}^{\frac{1}{4}}$
$(3 - {(1 - x^{7})}^{\frac{1}{4}}$
$3 - {(1 + x^{7})}^{\frac{1}{4}}$
$3 + {(1 + x^{7})}^{\frac{1}{4}}$

hint Hint:

We are given a function. We have to find it's inverse. To find an inverse, we will interchange the variables. We will replace x by y and y by x. Then we will find the equation in terms of y.

The correct answer is: $3 plus open parentheses 1 minus x to the power of 7 end exponent close parentheses to the power of fraction numerator 1 over denominator 4 end fraction end exponent$

The given function is y = $left parenthesis 1 minus open parentheses x minus 3 close parentheses to the power of 4 right square bracket to the power of 1 divided by 7 end exponent$
The first step will be to interchange the variables.
We get
$x equals left square bracket 1 minus open parentheses y minus 3 close parentheses to the power of 4 right square bracket to the power of 1 over 7 end exponent$
Now, we will find the above equation in terms of y.
$x equals left square bracket 1 minus open parentheses y minus 3 close parentheses to the power of 4 right square bracket to the power of 1 over 7 end exponent W e space w i l l space r a i s e space b o t h space t h e space s i d e s space t o space t h e space p o w e r space o f space 7 x to the power of 7 equals open square brackets 1 minus left parenthesis y minus 3 right parenthesis to the power of 4 close square brackets to the power of 1 over 7 cross times 7 end exponent x to the power of 7 equals 1 minus left parenthesis y minus 3 right parenthesis to the power of 4 R e a r r a n g i n g space t h e space e q u a t i o n left parenthesis y minus 3 right parenthesis to the power of 4 equals 1 minus x to the power of 7 W e space w i l l space t a k e space f o u r t h space r o o t space o f space b o t h space t h e space s i d e s left parenthesis y minus 3 right parenthesis to the power of 4 cross times 1 fourth end exponent equals left parenthesis 1 minus x to the power of 7 right parenthesis to the power of 1 fourth end exponent y minus 3 space equals left parenthesis 1 minus x to the power of 7 right parenthesis to the power of 1 fourth end exponent W e space w i l l space a d d space 3 space t o space b o t h space t h e space s i d e s. y space equals space 3 space plus space left parenthesis 1 space minus x to the power of 7 right parenthesis to the power of 1 fourth end exponent$
This is the required answer.

We used rules of indices to solve the question. For such questions, we should be familiar with rules of indices.

The domain of $f left parenthesis x right parenthesis equals square root of x to the power of 2 end exponent minus 4 vertical line x vertical line plus 3 end root$ is

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The inverse of the function is

We are given a function. We have to find it's inverse. To find an inverse, we will interchange the variables. We will replace x by y and y by x. Then we will find the equation in terms of y.

The correct answer is:

The given function is y = The first step will be to interchange the variables.We getNow, we will find the above equation in terms of y.This is the required answer.

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A Shell of mass M and radius R has a point mass placed at a distance r from its center. The gravitational potential energy u(r)-v will be,

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The correct graph representing the variation of total energy (E) kinetic energy and potential energy (U) of a satellite with its distance from the center of earth is.........

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The inverse of the function $y equals open square brackets 1 minus left parenthesis x minus 3 right parenthesis to the power of 4 end exponent close square brackets to the power of 1 divided by 7 end exponent$ is

The correct answer is: $3 plus open parentheses 1 minus x to the power of 7 end exponent close parentheses to the power of fraction numerator 1 over denominator 4 end fraction end exponent$

The domain of $f left parenthesis x right parenthesis equals square root of x to the power of 2 end exponent minus 4 vertical line x vertical line plus 3 end root$ is

The domain of $f left parenthesis x right parenthesis equals square root of x to the power of 2 end exponent minus 4 vertical line x vertical line plus 3 end root$ is

Calculate the d- and $l$ -isomers formed by the following compound.
$H O O C minus C H open parentheses C H subscript 3 end subscript close parentheses minus C H O H minus C H B r minus C H O H minus C H open parentheses C H subscript 3 end subscript close parentheses minus C O O H minus$

Calculate the d- and $l$ -isomers formed by the following compound.
$H O O C minus C H open parentheses C H subscript 3 end subscript close parentheses minus C H O H minus C H B r minus C H O H minus C H open parentheses C H subscript 3 end subscript close parentheses minus C O O H minus$