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Find the most general antiderivative of the function.

(Check your answers by differentiation.)

$ f(x) = \sqrt{2} $

$$\sqrt{2} x+C$$

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Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

So here we're trying to find the anti derivative of the function. Um and what it is is a five ax equals The Square Root of two. Before you think this is overly complicated, let's recognize what the square root of two is. The square root of two is a constant value. So this would be the same thing as if it's a f of X equals two, or f of X equals six. What we do is when we get F of X equals the square root of two, we know that the anti derivative of a constant, it's just a constant multiplied by X. So we have the square root of two times X. That right, there would be our anti derivative. And since we're producing the general anti derivative, we add the general constant C. This way, when we have F prime of X, regardless of what R. C is equal to, we see that our f prime of X is always going to be that this constant value of the square root of two. So based on that, we see that we've done this correctly.

California Baptist University