To find the inverse of the function \( f(x) = 3x^2 + 5 \), we will follow these steps:
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Replace \( f(x) \) with \( y \): \[ y = 3x^2 + 5 \]
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Swap \( x \) and \( y \): \[ x = 3y^2 + 5 \]
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Solve for \( y \):
- First, isolate the term with \( y^2 \): \[ x - 5 = 3y^2 \]
- Next, divide both sides by 3: \[ \frac{x - 5}{3} = y^2 \]
- Finally, take the square root of both sides: \[ y = \pm \sqrt{\frac{x - 5}{3}} \]
Since \( f(x) \) is not one-to-one over all real numbers (it is a parabola opening upwards), we generally restrict the domain to make it one-to-one. Let's assume we restrict \( f(x) \) to \( x \geq 0 \) to ensure \( f(x) \) is increasing. This means we only take the positive branch of the square root.
Thus, the inverse function is: \[ f^{-1}(x) = \sqrt{\frac{x - 5}{3}}, \quad \text{for } x \geq 5 \]
So the equation of the inverse is: \[ f^{-1}(x) = \sqrt{\frac{x - 5}{3}} \quad \text{for } x \geq 5 \]