Question

Given f(x) = 4x^4 find f^-1(x) Then state whether f^-1(x) is a function.

(This is precalculus, provide an explanation so I can hopefully not fail this class, thanks!)

Answer 1

• 
  `f^(-1)(x)=\pm\sqrt[4]{(x)/(4)}`
• 
  `f^(-1)(x) \quad\text{is not a function}`

Explanation:

To find the inverse function, solve for y:

`x=f(y)\\\\x=4y^4\\\\(x)/(4)=y^4\\\\\pm\sqrt[4]{(x)/(4)}=y\\\\f^(-1)(x)=\pm\sqrt[4]{(x)/(4)}`

f(x) is an even function, so f(-x) = f(x). Then the inverse relation is double-valued: for any given y, there can be either of two x-values that will give that result.

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A function is single-valued. That means any given domain value maps to exactly one range value. The test of this is the "vertical line test." If a vertical line intersects the graph in more than one point, then that x-value maps to more than one y-value.

The horizontal line test is similar. It is used to determine whether a function has an inverse function. If a horizontal line intersects the graph in more than one place, the inverse relation is not a function.

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Since the inverse relation for the given f(x) maps every x to two y-values, it is not a function. You can also tell this by the fact that f(x) is an even function, so does not pass the horizontal line test. When f(x) doesn't pass the horizontal line test, f^-1(x) cannot pass the vertical line test.

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The attached graph shows the inverse relation (called f₁(x)). It also shows a vertical line intersecting that graph in more than one place.