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Show that the function $g(x)=x^4+x^3+1$ is one-to-one on $[0,2]$.

My attempt

To prove one-to-oneness, we shall use the definition, that is, if $f(x_1)=f(x_2)$ , then $x_1=x_2$ for all $x_1,x_2\in[0,2]$

Suppose $f(x_1)=f(x_2)$, then ${x_1}^4+{x_1}^3+1={x_2}^4+{x_2}^3+1$

Which wasn't much of an attempt, as I got stuck.

Also I am rather new to proving conjections and surjections.

Help! Thanks in advance!

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Do you have any properties of one-to-one fuctions that you can use? For example, if $f$ and $g$ are both one-to-one, then $f\circ g$ is one-to-one. Can you use that? –  Todd Wilcox Oct 29 '12 at 15:12
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Hint: derivative. –  Did Oct 29 '12 at 15:13
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2 Answers 2

up vote 2 down vote accepted

The derivative is $x \mapsto 4x^3 + 3x^2$ which is non-negative on $[0,2]$. So your functions is increasing on $[0,2]$ so it is one-to-one on $[0,2]$.

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Not positive at 0. –  Did Oct 29 '12 at 15:14
    
I meant non-negative. In my country (France), x positive <=> x >= 0 and x strictly positive <=> x > 0 –  xavierm02 Oct 29 '12 at 15:16
    
Does it work the same for increasing? Should I put "non-decreasing"? –  xavierm02 Oct 29 '12 at 15:17
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The function is increasing on $[0,2]$ but the derivative is $>0$ only on $(0,2]$. Hence more care is needed than in the present version of this answer. –  Did Oct 29 '12 at 15:32
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First, it would be easier to just prove that the function is increasing on the interval. That being said, we may be able to push your solution a bit further. Using difference of fourths and difference of cubes, we have $$0=x_1^4 - x_2^4 + x_1^3 - x_2^3$$ $$=(x_1 - x_2)(x_2^3 + x_2^2x_1 + x_2x_1^2 + x_1^3) + (x_1 - x_2)(x_2^2 + x_1x_2 + x_1^2)$$ factoring out $(x_1 - x_2)$, it's easy to see that the latter part will be positive on the interval, this requires $x_1 = x_2$.

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Why are you allowed to factor out $x_1-x_2$ if they're equal? –  VF1 Oct 29 '12 at 15:17
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@VF1 $xy + xz = x(y + z)$ –  EuYu Oct 29 '12 at 15:19
    
Oh, sorry, never mind. For some reason I interpreted "factor" as "divide out by." My mistake. –  VF1 Oct 29 '12 at 15:20
    
Well @VF1, that's exactly what was meant by "factor out" in the above answer: to divide out by! Of course we can assume $\,x_1\neq x_2\,$ , after all we're trying to prove injectivity –  DonAntonio Oct 29 '12 at 16:56
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