Question

Find inverse of this
pls ​

Answer 1

GiveN :-

• A function is given to us.
• $\sf y= \sqrt{ (x-1)(7-x)}$

To FinD :-

• The inverse of the function .

SolutioN :-

A function is given to us and we need to find its Inverse . The given function is ,

$\bf \to y = \sqrt{ (x-1)(7-x)}$

For it Firstly Interchange the variables . The equation becomes ,

$\sf :\pink{\implies x =\sqrt{ (y-1)(7-y)}}$

Now secondly solve for y , we have ,

$\sf :\implies x^2=(\sqrt{(y-1)(7-y)})^2$

$\sf:\implies x^2= (y-1)(7-y) \\\\\sf:\implies x^2=y(7-y)-1(7-y) \\\\\sf:\implies x^2= 7y - y^2 -7+y \\\\\sf:\implies -x^2 -y^2 +8y -7 = 0 \\\\\sf:\implies -y^2+8y -x^2 - 7 = 0$

Now considering this equation as a quadratic equation , we can find the roots using Shreedhacharya's Formula ( Quadratic Formula ) . With respect to Standard form here , a = (-1) , b = 8 and c = -x² - 7 . Now substituting all the values in the Quadratic Formula , we have ,

$\sf \to y = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\\\\sf\to y =\dfrac{ -8\pm \sqrt{ 8^2-4(-1)( -x^2-7)}}{2(-1)}\\\\\sf\to y = \dfrac{-8\pm \sqrt{ 64 -4(x^2+7)}}{-2} \\\\\sf\to y =\dfrac{-8\pm 2\sqrt{16-x^2-7}}{-2}\\\\\sf\to y =\dfrac{-8\pm 2\sqrt{-x^2+9}}{-2}\\\\\sf\to y =\dfrac{-8 \pm 2\sqrt{(3-x)(3+x)}}{-2} \\\\\sf:\implies \boxed{\pink{\sf y = 4 \pm \sqrt{(3+x)(3-x)} }}$

Now let's replace y with f-¹(x) . So that ,

$\bf \to\boxed{\red{ \bf f^{-1}(x)= 4 \pm \sqrt{(3+x)(3-x)}}}$

$\rule{200}2$

Now let's check . If the domain of f-¹(x) is equal to the range of f(x) , then our answer will be correct . The range is the set of all valid y values. Using the graph we see that the Range of f(x) is [ 0 , 3 ]

$\to \boxed{\bf \mathbb{R} \in [ 0 , 3 ]}$

Now let's find the domain of f-¹(x) . For that , the value inside the squareroot must be Greater than or equal to zero . That is ,

$\bf\to (3+x)(3-x) \geq 0 \\\\\bf\to \boxed{\bf x \in [ -3 , 3] }$

Now here we see that the domain of f-¹(x) is not equal to the range of f(x) . Therefore there is no Inverse of that function .

$\qquad\qquad\underset{\blue{\sf \therefore No \ Inv erse}}{\underbrace{\boxed{\pink{\bf{ Domain \neq Range }}}}}$