given a function f(x), how do we evaluate f^-1(4)?

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Given a function f(x), how do we evaluate f^-1(4)? This question isn't hard but it could still be tricky for many students. Subscribe to @bprpmathbasics for more algebra and precalculus tutorials.

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If I need to get a lot of solutions of this form quickly, I might use the first method. But, for displaying the solution to others, or to myself to check my own work, I'd use the second method. They're pretty much equivalent but have different levels of explanatory power. You got to the solution faster with the first method of course, but with the second method I fully understood what you were doing whether I had previous learning on this or not.

Tletna
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Through the years I've often stumbled on this very simple concept, partly because of notation. Effectively we're drawing a diagonal line on the 2D graph plane, and reflecting the function by folding on that 45 degree line. So, if f(x) is e^x, then f-1(x) is ln(x). And if y=xe^(x), then the inverse is W(x). You give me y, I'll give you x, it's all about the input data.

spelunkerd
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The first method is the easiest and fastest, while the second is the most complete, 'cause we find the equation of the inverse function.

leandro
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Thank you, some book report only one method ✌🏻✌🏻

albertow.
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I'll admit it always confuses me that we say both f(x) and f^-1(x), I find it easier to consider f(x) and f^-1(y), or at least use two different variables and solve for the one we care about in both cases; perhaps it's because I'm more on the physics side of things, where units matter and we're used to solve for things that aren't called x, but I find it easy to mess up and confuse which x it is in which context otherwise.

puissantpoisson
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Stop giving easy algebra questions. You're making me feel like i'm good at algebra

Twilight