Find the Inverse f(x)=3x-4+2x^2

$f (x) = 3 x - 4 + 2 x^{2}$

Replace $f (x)$ with $y$ .

$y = 3 x - 4 + 2 x^{2}$

Interchange the variables.

$x = 3 y - 4 + 2 y^{2}$

Solve for $y$ .

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Rewrite the equation as $3 y - 4 + 2 y^{2} = x$ .

$3 y - 4 + 2 y^{2} = x$

Move $x$ to the left side of the equation by subtracting it from both sides.

$3 y - 4 + 2 y^{2} - x = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = 3$ , and $c = - 4 - x$ into the quadratic formula and solve for $y$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (2 \cdot (- 4 - x))}}{2 \cdot 2}$

Simplify.

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Simplify the numerator.

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Raise $3$ to the power of $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (2 \cdot (- 4 - x))}}{2 \cdot 2}$

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (2 \cdot - 4 + 2 (- x))}}{2 \cdot 2}$

Multiply $2$ by $- 4$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (- 8 + 2 (- x))}}{2 \cdot 2}$

Multiply $- 1$ by $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (- 8 - 2 x)}}{2 \cdot 2}$

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 8 - 4 (- 2 x)}}{2 \cdot 2}$

Multiply $- 4$ by $- 8$ .

$y = \frac{- 3 \pm \sqrt{9 + 32 - 4 (- 2 x)}}{2 \cdot 2}$

Multiply $- 2$ by $- 4$ .

$y = \frac{- 3 \pm \sqrt{9 + 32 + 8 x}}{2 \cdot 2}$

Add $9$ and $32$ .

$y = \frac{- 3 \pm \sqrt{41 + 8 x}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$y = \frac{- 3 \pm \sqrt{41 + 8 x}}{4}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $3$ to the power of $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (2 \cdot (- 4 - x))}}{2 \cdot 2}$

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (2 \cdot - 4 + 2 (- x))}}{2 \cdot 2}$

Multiply $2$ by $- 4$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (- 8 + 2 (- x))}}{2 \cdot 2}$

Multiply $- 1$ by $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (- 8 - 2 x)}}{2 \cdot 2}$

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 8 - 4 (- 2 x)}}{2 \cdot 2}$

Multiply $- 4$ by $- 8$ .

$y = \frac{- 3 \pm \sqrt{9 + 32 - 4 (- 2 x)}}{2 \cdot 2}$

Multiply $- 2$ by $- 4$ .

$y = \frac{- 3 \pm \sqrt{9 + 32 + 8 x}}{2 \cdot 2}$

Add $9$ and $32$ .

$y = \frac{- 3 \pm \sqrt{41 + 8 x}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$y = \frac{- 3 \pm \sqrt{41 + 8 x}}{4}$

Change the $\pm$ to $+$ .

$y = \frac{- 3 + \sqrt{41 + 8 x}}{4}$

Rewrite $- 3$ as $- 1 (3)$ .

$y = \frac{- 1 \cdot 3 + \sqrt{41 + 8 x}}{4}$

Factor $- 1$ out of $\sqrt{41 + 8 x}$ .

$y = \frac{- 1 \cdot 3 - 1 (- \sqrt{41 + 8 x})}{4}$

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{41 + 8 x})$ .

$y = \frac{- 1 (3 - \sqrt{41 + 8 x})}{4}$

Move the negative in front of the fraction.

$y = - \frac{3 - \sqrt{41 + 8 x}}{4}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $3$ to the power of $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (2 \cdot (- 4 - x))}}{2 \cdot 2}$

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (2 \cdot - 4 + 2 (- x))}}{2 \cdot 2}$

Multiply $2$ by $- 4$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (- 8 + 2 (- x))}}{2 \cdot 2}$

Multiply $- 1$ by $2$ .

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot (- 8 - 2 x)}}{2 \cdot 2}$

Apply the distributive property.

$y = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 8 - 4 (- 2 x)}}{2 \cdot 2}$

Multiply $- 4$ by $- 8$ .

$y = \frac{- 3 \pm \sqrt{9 + 32 - 4 (- 2 x)}}{2 \cdot 2}$

Multiply $- 2$ by $- 4$ .

$y = \frac{- 3 \pm \sqrt{9 + 32 + 8 x}}{2 \cdot 2}$

Add $9$ and $32$ .

$y = \frac{- 3 \pm \sqrt{41 + 8 x}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$y = \frac{- 3 \pm \sqrt{41 + 8 x}}{4}$

Change the $\pm$ to $-$ .

$y = \frac{- 3 - \sqrt{41 + 8 x}}{4}$

Factor $- 1$ out of $- 3 - \sqrt{41 + 8 x}$ .

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Reorder $41$ and $8 x$ .

$y = \frac{- 3 - \sqrt{8 x + 41}}{4}$

Rewrite $- 3$ as $- 1 (3)$ .

$y = \frac{- 1 \cdot 3 - \sqrt{8 x + 41}}{4}$

Factor $- 1$ out of $- \sqrt{8 x + 41}$ .

$y = \frac{- 1 \cdot 3 - (\sqrt{8 x + 41})}{4}$

Factor $- 1$ out of $- 1 (3) - (\sqrt{8 x + 41})$ .

$y = \frac{- 1 (3 + \sqrt{8 x + 41})}{4}$

Rewrite $- 1 (3 + \sqrt{8 x + 41})$ as $- (3 + \sqrt{8 x + 41})$ .

$y = \frac{- (3 + \sqrt{8 x + 41})}{4}$

Move the negative in front of the fraction.

$y = - \frac{3 + \sqrt{8 x + 41}}{4}$

The final answer is the combination of both solutions.

$y = - \frac{3 - \sqrt{41 + 8 x}}{4}$

$y = - \frac{3 + \sqrt{8 x + 41}}{4}$

$y = - \frac{3 - \sqrt{41 + 8 x}}{4}$

$y = - \frac{3 + \sqrt{8 x + 41}}{4}$

Solve for $y$ and replace with $f^{- 1} (x)$ .

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Replace the $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = - \frac{3 - \sqrt{41 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 41}}{4}$

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{41 + 8 (3 x - 4 + 2 x^{2})}}{4}$

Simplify the numerator.

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Apply the distributive property.

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{41 + 8 (3 x) + 8 \cdot - 4 + 8 (2 x^{2})}}{4}$

Simplify.

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Multiply $3$ by $8$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{41 + 24 x + 8 \cdot - 4 + 8 (2 x^{2})}}{4}$

Multiply $8$ by $- 4$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{41 + 24 x - 32 + 8 (2 x^{2})}}{4}$

Multiply $2$ by $8$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{41 + 24 x - 32 + 16 x^{2}}}{4}$

Subtract $32$ from $41$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{24 x + 9 + 16 x^{2}}}{4}$

Reorder terms.

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{16 x^{2} + 24 x + 9}}{4}$

Factor using the perfect square rule.

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Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{{(4 x)}^{2} + 24 x + 9}}{4}$

Rewrite $9$ as $3^{2}$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{{(4 x)}^{2} + 24 x + 3^{2}}}{4}$

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot (4 x) \cdot 3$

Simplify.

$2 a b = 24 x$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 4 x$ and $b = 3$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - \sqrt{{(4 x + 3)}^{2}}}{4}$

Pull terms out from under the radical, assuming positive real numbers.

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - (4 x + 3)}{4}$

Apply the distributive property.

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - (4 x) - 1 \cdot 3}{4}$

Multiply $4$ by $- 1$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - 4 x - 1 \cdot 3}{4}$

Multiply $- 1$ by $3$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{3 - 4 x - 3}{4}$

Subtract $3$ from $3$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{- 4 x + 0}{4}$

Add $- 4 x$ and $0$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{- 4 x}{4}$

Cancel the common factor of $- 4$ and $4$ .

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Factor $4$ out of $- 4 x$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{4 (- x)}{4}$

Cancel the common factors.

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Factor $4$ out of $4$ .

$g (3 x - 4 + 2 x^{2}) = - \frac{4 (- x)}{4 (1)}$

Cancel the common factor.

$g (3 x - 4 + 2 x^{2}) = - \frac{4 (- x)}{4 \cdot 1}$

Rewrite the expression.

$g (3 x - 4 + 2 x^{2}) = - \frac{- x}{1}$

Divide $- x$ by $1$ .

$g (3 x - 4 + 2 x^{2}) = x$

Since $g (f (x)) = x$ , $f^{- 1} (x) = - \frac{3 - \sqrt{41 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 41}}{4}$ is the inverse of $f (x) = 3 x - 4 + 2 x^{2}$ .

$f^{- 1} (x) = - \frac{3 - \sqrt{41 + 8 x}}{4}, - \frac{3 + \sqrt{8 x + 41}}{4}$

Find the Inverse f(x)=3x-4+2x^2

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