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f(x) = x^4 + 1, x ? 0 Find an explicit formula for f?¹. f?¹(x) = Graph f?¹, f, and the line y = x on the same screen. To check your work, see whether the graphs of f and f?¹ are reflections about the line.

          f(x) = x^4 + 1, x ? 0
Find an explicit formula for f?¹.
f?¹(x) =
Graph f?¹, f, and the line y = x on the same screen. To check your work, see whether the graphs of f and f?¹ are reflections about the line.
        
f(x) = x^4 + 1, x ? 0
Find an explicit formula for f?¹.
f?¹(x) =
Graph f?¹, f, and the line y = x on the same screen. To check your work, see whether the graphs of f and f?¹ are reflections about the line.

Added by Shawn J.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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f(x) = x^4 + 1, x ≥ 0 Find an explicit formula for f⁻¹. f⁻¹(x) = Graph f⁻¹, f, and the line y = x on the same screen. To check your work, see whether the graphs of f and f⁻¹ are reflections about the line.
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Transcript

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00:01 I actually think it might be easier to tell you the answer on the graph first, because you have x to the fourth plus one where x is greater than or equal to zero.
00:11 So if you understand your transformations, this graph is shifted up one, and we're only on the right side of the origin.
00:19 So as i'm looking at the graphs, in x to the fourth, it shoots up in the graph.
00:27 So it starts at 0 -1 and shoots up, and that's the graph of f.
00:33 And the only one that does that is the bottom right graph.
00:38 So i was going to write that in green.
00:41 The bottom right.
00:46 And its inverse would then be reflected over the line, as i mentioned, and it shoots to the right then.
00:52 And that's f inverse.
00:54 So let me go ahead and circle that.
00:56 Now the actual way to find the inverse, what i like to do is switch the x and the y coordinates around and then solve for y...
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