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Let $f:[0,1] \to [0,2]$ be given by $f(x)=x^2+x^4$. Determine if f is an invertible function. If it is, compute its inverse. If it is not invertible, provide an argument to support your claim.

I think that $f$ is invertible because the restrictions on the domain and codomain make it bijective, but I am struggling to compute the inverse. Usually, if I was given $y=f(x)$ I would just rearrange to make x the subject and then swap $x$ and $y$, but I don't know how to do that for this example. Any help would be greatly appreciated.

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  • $\begingroup$ Completing square $\endgroup$ – BAI Oct 28 '17 at 13:01
  • $\begingroup$ Setting $y=x^2+x^4$ and solve this equation and Substitute $x^2=t$ $\endgroup$ – Dr. Sonnhard Graubner Oct 28 '17 at 13:04
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$y=f(x)=x^2+x^4$ $$y=(x^2+1/2)^2-1/4$$ $$\sqrt{y+1/4}-1/2=x^2$$ $$\sqrt{\sqrt{y+1/4}-1/2}=x$$ The last step is valid since $x,y\ge0$. Hence, $$f^{-1}(y)= \sqrt{\sqrt{y+1/4}-1/2}=x $$

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  • $\begingroup$ Thankyou! I never would have thought to complete the square. $\endgroup$ – T.meeley Oct 28 '17 at 14:29
  • $\begingroup$ @T.meeley you may accept or upvote this answer if you think it helps you :) $\endgroup$ – BAI Oct 29 '17 at 6:37
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Hint: $f'(x)=2x+4x^3\geq0$ it is only zero in $x=0$ on the given domain, hence the function is monotonic and invertible.

Now, to the inversion:

$$f(x)=x^4+x^2 \implies x^4+x^2-f(x)=0 \text{ use quadratic formula }$$ $$\implies x^2_{1,2}=\dfrac{-1\pm\sqrt{1+4f(x)}}{2}.$$

Can you complete it from here? Note that you can only take the square-root for the positive case of $\pm$.

$$\implies x_{3,4}=\pm\sqrt{\dfrac{-1+\sqrt{1+4f(x)}}{2}}.$$

Now use the domain and the image of the function from the beginning to rule out which sign is correct.

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