## JEE mains tips and tricks for Inverse Function JEE mains tips and tricks : What is a function?

A function of real variables, f:A→B, is a relationship between two sets of real numbers A and B, that every number x of A corresponds to one, and only one, number of B, that relationship is called f(x) and calls the image of x using f.

From this moment forward, we will assume f:A→B, being A and B subsets of the real number set R. A is the domain of f and B it’s its codomain, also known as image .

Note: B is the codomain and follows f(A)⊆B. The set f(A) is the image of f. Informally, the inverse function of f is the function $f^{-1}$:B→A in the way that given a number y of B, it allows to find the number x of A in which y=f(x). For this, you write $f^{-1}\left ( y \right )$

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## Example:

If f(x)=2x, it’s inverse is $f^{-1}$(x)=x/2. To check if this is true, let’s use the number 8. $f^{-1}$(8)=8/2=4

Indeed, the image of 4 is 8:

f(4)=2⋅4=8

## Things to consider

The function f(x)= $x^{2}$

doesn’t have an inverse function. We can understand this easily with an example, f−1(4) could be $f^{-1}$(4)=2 or it also could be $f^{-1}$(4)=−2: Note: if we restrict the dominion of f to the negative real numbers or positive real numbers then the function does have an inverse.

Taking into account the definition given for the function f, as $f^{-1}$ is also a function, we must demand that each number y of B has a unique image x= $f^{-1}$(y) in A. The number x that comply $f^{-1}$(y)=x it’s called anti-image of y.

Mathematically, this requirement of uniqueness of the anti-image (in order to be a function) it’s translated demanding that the function f it’s injective:

The function f:A→B is injective if it fulfills:

f(x)=f(y)↔x=y,∀x,y∈A

Meaning, f is injective if: the image of two numbers of A are equals only, and just only, said numbers of A are the same number.

Also, as a function, the inverse of $f^{-1}$ must give the anti-image of all the numbers of B. Nevertheless, this is not possible if some number of B is not the image of a number of A. Mathematically, this problem is solved demanding that the function f is surjective:

The function f:A→B is surjective if it fulfills:

∀y∈B, ∃x∈A, f(x)=y

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Meaning that, f is surjective if every number in B is the image through f of some number in A.

Due to the importance of simultaneous injectivity and surjectivity of a function, there is a definition that considers both properties:

The function f:A→B is bijective if it is injective and surjective.

Definition of inverse function

We will now see the formal definition of inverse function:

The inverse function $f^{-1}$

of a bijective function f:A→B is the function $f^{-1}$:B→A which complies: $f^{-1}$(f(x))=x, ∀x∈A

f( $f^{-1}$(y))=y, ∀y∈B

Another way of saying it: $f^{-1}$∘f=idA

f∘ $f^{-1}$=idB

where idA is the identity function of A:

idA:A→A

idA(x)=x, ∀x∈A

and idB is the identity function of B:

idB:B→B

idB(y)=y, ∀y∈B

#### Issue 1

If $f^{-1}$:B→A complies with the conditions given in the definition of inverse function of f, is it really the inverse function of f?

Meaning, $f^{-1}$

gives the anti-image of f?

Let us suppose that x is a number of A. Then, it has an image through f:∃y=f(x)∈B

In fact, $f^{-1}$

gives the anti-image of y applying the first condition: $f^{-1}$(y)= $f^{-1}$(f(x))=x

#### Issue 2

Is the inverse function unique?

The answer is yes, let us see the reasoning:

Let us suppose that the function g:B→A complies with the conditions of the inverse function of f.

If y is a number of B, as f is a bijective function, we can assure that a number x of A exist that is y = f(x). You get that:

g(y)=x

And also: $f^{-1}$(y)=x

Meaning that:

g(y)= $f^{-1}$(y), ∀y∈B

This demonstrates that the inverse is unique, because the image of any number y of B through g is the same as the image through $f^{-1}$.

Meaning that the functions are the same, since they are defined between the same sets of numbers, and the image of every number of B coincides.

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## Calculating inverse function

So, for a function to be an inverse function $f^{-1}$ of a function f it must comply:

f(a) = b $f^{-1}$(b) = a

Let’s use the example of the function f(x) = x + 4 We can see that:The dominion of $f^{-1}$ is the image of f.

The image of $f^{-1}$ is the dominion of f.

If we want to find the image of a function we have to find the dominion of it’s inverse function.

If two functions are inverse functions, it’s composition is the identity function.

(f o $f^{-1}$) (x) = ( $f^{-1}$

${f}^{–1}$

o f) (x) = x

The graphs of f and $f^{-1}$ are symmetrical with respect to the first and third quadrant bisector. We must distinguish between an inverse function , $f^{-1}$, and the inverse of function:

$\frac{1/}{\mathrm{f\left(}\left(\mathrm{x\right)}\right)}$

.

The inverse of the function f(x) = x + 4 is

$\frac{1/\left(}{x+4\right)}$

The inverse function of f(x) = x + 4 is $f^{-1}$(x) = x – 4 because the composition of the two functions is the identity function:

g ∘ f = g[f(x)] = g(x + 4) = x + 4 – 4 = x

## How do you calculate the inverse function?

##### 3. Interchange the variables.

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## Example 1.

Calculate the inverse function of:

f(x)= $\frac{2x+3}{x-1}$

1. First change f(x) for y and move the denominator to the other side of the equation:

y= $\frac{2x+3}{x-1}$

y(x-1) = 2x+3

Then is just simple math

$yx–y=2x+3$

$\phantom{\rule{0ex}{0ex}}yx–2x=y+3$

To clear the x

$x\left(y–2\right)=y+3$

$\phantom{\rule{0ex}{0ex}}x=\left(\frac{y+3\right)/\left(}{y–2\right)}$

Finally, we change x for $f^{-1}$(x) and y for x:

$\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left[/extract_itex\right]\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left[/extract_itex\right]\left(x\right)=\frac{x+3}{x–2}$

Now let’s check the result for x = 2

= 7

${\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left[/extract_itex\right]\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left[/extract_itex\right]\left(}^{}\left(7\right)\right)=\left(\frac{7+3\right)/\left(}{7–2\right)}=2$

## Example 2. $f(x)=\sqrt{x-1}$

We replace f(x) for y $y=\sqrt{x-1}$

After that, it’s just simple math $y^{3}=x-1$ $x=y^{3}+1$

Finally, we clear the x and change x for $f^{-1}$(x), and the y for x

$\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D%28x%29%3Dx%5E%7B3%7D%2B1&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left(x\right)=x^\left\{3\right\}+1\left[/extract_itex\right]\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D%28x%29%3Dx%5E%7B3%7D%2B1&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left(x\right)=x^\left\{3\right\}+1\left[/extract_itex\right]$

## Example 3. $f(x)=x^{2}$

We replace f(x) for y

After that, it’s just simple math $y=x^{2}$

x= $\pm \sqrt{y}$

$\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left[/extract_itex\right]\left[Image_Link\right]https://s0.wp.com/latex.php?latex=f%5E%7B-1%7D&bg=ffffff&fg=111111&s=0&c=20201002 \left[/Image_Link\right] \left[extract_itex\right]f^\left\{-1\right\}\left[/extract_itex\right]=±\sqrt{x}$

But wait, this is not a function!

The inverse function doesn’t exist because every element in the domain has two images and function only has one image per element in the domain.

If you want to read more about Inverse Function you can buy Differential Calculus By S.K. Goyal.

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