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How do you find the inverse of h(x) = x + x^0.5?
I understand how to switch x and y, but how can we isolate the y?
 2 years ago
 2 years ago
How do you find the inverse of h(x) = x + x^0.5? I understand how to switch x and y, but how can we isolate the y?
 2 years ago
 2 years ago

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malevolence19Best ResponseYou've already chosen the best response.1
The inverse is: \[\frac{1}{2}(2y1)\pm \frac{1}{2}\sqrt{4y+1}\]
 2 years ago

TransendentialPIBest ResponseYou've already chosen the best response.0
Thanks. Did you use software to get this? I'm needing to do this by hand. h(x) looks to be one to one and looking at the derivatives it looks like h(x) is one to one. I'll keep looking, thanks.
 2 years ago

TransendentialPIBest ResponseYou've already chosen the best response.0
I was helping someone with this exercise. They left out the fact we were looking for \[h ^{1}(6)\] given \[h(x)=x+\sqrt x\] Here is what I came up with: This is asking when does h(x) = 6 6 = x + x^.5 0= x + x^.5 6 factors to 0=(x^.5 + 3)(x^.5  2) x = 9 and x = 4 Checking in the original and by looking at the graph of h, we see the only place h(x) = 6 is at x = 4. h1(6)=4 Maybe this will help someone some time.
 2 years ago
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