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I came across this question and I am not sure on how to go about solving this question. Any advice or hints would be appreciated.

The question is: Let $j$ and $n$ be integers such that $1\leqslant j\leqslant n$. Let $g_{jn} (x) =(1+jx)(1-x)^n $ be a function defined $0\leqslant x\leqslant 1$. Show that for all positive integers $j\leqslant n$, the function is a bijective function and the range of the function is $0\leqslant x\leqslant 1$.

Any ideas on how to start with this question?

Maths_noob_123
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1 Answers1

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$$g_{jn} ' (x) =j(1-x)^n - n(1-x)^{n-1} (1+jx) =(1-x)^{n-1} ((j-n)-jx(1+n))<0$$ thus the function is monotone decreasing. Moreover $$g_{jn} (0)=1 , g_{jn}(1)=0 $$ thus the image $$g_{jn} ([0,1])=[0,1].$$

  • Thank you! I have a much better understanding of this now! Thank you for helping me edit the question too! Really appreciated it! – Maths_noob_123 Sep 16 '16 at 13:22