INVERSE FUNCTIONS $\newline$ rmula for the inverse of the function. $\newline$ $\sqrt{2+3 x}$ $\newline$ $22$ . $f(x)=\frac{4 x-1}{2 x+3}$ $\newline$ $24$ . $y=x^{2}-x, \quad x \geqslant \frac{1}{2}$

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Q. INVERSE FUNCTIONS

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rmula for the inverse of the function.

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\sqrt{2+3 x}

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22

f(x)=\frac{4 x-1}{2 x+3}

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24

y=x^{2}-x, \quad x \geqslant \frac{1}{2}

Find Inverse of $\sqrt{2+3x}$ : First, let's find the inverse of $f(x) = \sqrt{2+3x}$ . Swap $f(x)$ with $y$ : $y = \sqrt{2+3x}$ . Now, swap $x$ and $y$ to find the inverse: $x = \sqrt{2+3y}$ . Square both sides to get rid of the square root: $x^2 = 2 + 3y$ .
Solve for Inverse: Solve for $y$ : $3y = x^2 - 2$ . $\newline$ Divide by $3$ : $y = \frac{x^2 - 2}{3}$ . $\newline$ This is the inverse function of $f(x) = \sqrt{2+3x}$ .
Quadratic Formula Solution: Now, let's find the inverse of $f(x) = \frac{4x-1}{2x+3}$ . $\newline$ Swap $f(x)$ with $y$ : $y = \frac{4x-1}{2x+3}$ . $\newline$ Swap $x$ and $y$ to find the inverse: $x = \frac{4y-1}{2y+3}$ . $\newline$ Multiply both sides by $(2y+3)$ to get rid of the denominator: $x(2y+3) = 4y - 1$ .
Quadratic Formula Solution: Now, let's find the inverse of $f(x) = \frac{4x-1}{2x+3}$ . Swap $f(x)$ with $y$ : $y = \frac{4x-1}{2x+3}$ . Swap $x$ and $y$ to find the inverse: $x = \frac{4y-1}{2y+3}$ . Multiply both sides by $(2y+3)$ to get rid of the denominator: $x(2y+3) = 4y - 1$ . Expand the left side: $2xy + 3x = 4y - 1$ . Move all $y$ terms to one side: $f(x)$ $1$ . Factor out $y$ : $f(x)$ $3$ .
Quadratic Formula Solution: Now, let's find the inverse of $f(x) = \frac{4x-1}{2x+3}$ .
Swap $f(x)$ with $y$ : $y = \frac{4x-1}{2x+3}$ .
Swap $x$ and $y$ to find the inverse: $x = \frac{4y-1}{2y+3}$ .
Multiply both sides by $(2y+3)$ to get rid of the denominator: $x(2y+3) = 4y - 1$ .Expand the left side: $2xy + 3x = 4y - 1$ .
Move all $y$ terms to one side: $f(x)$ $1$ .
Factor out $y$ : $f(x)$ $3$ .Divide by $f(x)$ $4$ to solve for $y$ : $f(x)$ $6$ .
This is the inverse function of $f(x) = \frac{4x-1}{2x+3}$ .
Quadratic Formula Solution: Now, let's find the inverse of $f(x) = \frac{4x-1}{2x+3}$ . Swap $f(x)$ with $y$ : $y = \frac{4x-1}{2x+3}$ . Swap $x$ and $y$ to find the inverse: $x = \frac{4y-1}{2y+3}$ . Multiply both sides by $(2y+3)$ to get rid of the denominator: $x(2y+3) = 4y - 1$ . Expand the left side: $2xy + 3x = 4y - 1$ . Move all $y$ terms to one side: $f(x)$ $1$ . Factor out $y$ : $f(x)$ $3$ . Divide by $f(x)$ $4$ to solve for $y$ : $f(x)$ $6$ . This is the inverse function of $f(x) = \frac{4x-1}{2x+3}$ . Lastly, let's find the inverse of $f(x)$ $8$ , where $f(x)$ $9$ . Swap $y$ with $x$ : $y$ $2$ . Now we need to solve for $y$ , but this is a quadratic equation, so we'll use the quadratic formula. The quadratic formula is $y$ $4$ , where $y$ $5$ .
Quadratic Formula Solution: Now, let's find the inverse of $f(x) = \frac{4x-1}{2x+3}$ . Swap $f(x)$ with $y$ : $y = \frac{4x-1}{2x+3}$ . Swap $x$ and $y$ to find the inverse: $x = \frac{4y-1}{2y+3}$ . Multiply both sides by $(2y+3)$ to get rid of the denominator: $x(2y+3) = 4y - 1$ . Expand the left side: $2xy + 3x = 4y - 1$ . Move all $y$ terms to one side: $f(x)$ $1$ . Factor out $y$ : $f(x)$ $3$ . Divide by $f(x)$ $4$ to solve for $y$ : $f(x)$ $6$ . This is the inverse function of $f(x) = \frac{4x-1}{2x+3}$ . Lastly, let's find the inverse of $f(x)$ $8$ , where $f(x)$ $9$ . Swap $y$ with $x$ : $y$ $2$ . Now we need to solve for $y$ , but this is a quadratic equation, so we'll use the quadratic formula. The quadratic formula is $y$ $4$ , where $y$ $5$ . In our equation, $y$ $6$ , $y$ $7$ , and $y$ $8$ . Plug these into the quadratic formula: $y$ $9$ . Simplify: $y = \frac{4x-1}{2x+3}$ $0$ .
Quadratic Formula Solution: Now, let's find the inverse of $f(x) = \frac{4x-1}{2x+3}$ . Swap $f(x)$ with $y$ : $y = \frac{4x-1}{2x+3}$ . Swap $x$ and $y$ to find the inverse: $x = \frac{4y-1}{2y+3}$ . Multiply both sides by $(2y+3)$ to get rid of the denominator: $x(2y+3) = 4y - 1$ . Expand the left side: $2xy + 3x = 4y - 1$ . Move all $y$ terms to one side: $f(x)$ $1$ . Factor out $y$ : $f(x)$ $3$ . Divide by $f(x)$ $4$ to solve for $y$ : $f(x)$ $6$ . This is the inverse function of $f(x) = \frac{4x-1}{2x+3}$ . Lastly, let's find the inverse of $f(x)$ $8$ , where $f(x)$ $9$ . Swap $y$ with $x$ : $y$ $2$ . Now we need to solve for $y$ , but this is a quadratic equation, so we'll use the quadratic formula. The quadratic formula is $y$ $4$ , where $y$ $5$ . In our equation, $y$ $6$ , $y$ $7$ , and $y$ $8$ . Plug these into the quadratic formula: $y$ $9$ . Simplify: $y = \frac{4x-1}{2x+3}$ $0$ . Since $f(x)$ $9$ , we take the positive square root to ensure $y$ is also greater than or equal to $y = \frac{4x-1}{2x+3}$ $3$ . So, the inverse function is $y = \frac{4x-1}{2x+3}$ $4$ .

INVERSE FUNCTIONS\newlinermula for the inverse of the function.\newline2+3x \sqrt{2+3 x} 2+3x​\newline222222. f(x)=4x−12x+3 f(x)=\frac{4 x-1}{2 x+3} f(x)=2x+34x−1​\newline242424. y=x2−x,x⩾12 y=x^{2}-x, \quad x \geqslant \frac{1}{2} y=x2−x,x⩾21​

INVERSE FUNCTIONS $\newline$ rmula for the inverse of the function. $\newline$ $\sqrt{2+3 x}$ $\newline$ $22$ . $f(x)=\frac{4 x-1}{2 x+3}$ $\newline$ $24$ . $y=x^{2}-x, \quad x \geqslant \frac{1}{2}$