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TransendentialPI

  • 4 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?

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  1. malevolence19
    • 4 years ago
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    The inverse is: \[\frac{1}{2}(-2y-1)\pm \frac{1}{2}\sqrt{4y+1}\]

  2. TransendentialPI
    • 4 years ago
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    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.

  3. TransendentialPI
    • 4 years ago
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    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. h-1(6)=4 Maybe this will help someone some time.

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