f:R->R be such that f(x)=x^3-3x^2+5x -10..prove that whether it is one to one or onto or both..

Question

please give me adetail solution.basically I am getting confused how to determine onto function or not for a polynomial.please answer me as early as possible..

Answer 1

$f(x) = x^3 - 3x^2 +5x - 10$

$f'(x) = 3x^2 - 6x + 5$

Now, a polynomial is bijective (one-one and onto) if and only if its derivative never changes sign.

Here, the derivative is $3x^2 - 6x + 5$, for any $x$, this is positive. So, $f$ is one-one and onto. 

Suppose $f$ is not bijective. Then we can do as follows:

$f'(x) = 0 \implies x = 6 \pm \frac{\sqrt{36 - 60}} {6}$, hence no real roots. 

Now, all polynomial functions are continuous (Ref: http://www.themathpage.com/aCalc/continuous-function.htm). So, without any maximum or minimum, the given function must be one-one. 

Comments

derivative changes its sign means sometimes it will be positive,sometimes negetive???

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sir please tell me about the second part ,giving unreal roots..

Suppose f is not bijective. Then we can do as follows:

f′(x)=0⟹x=6±36−60√6, hence no real roots. 

Now, all polynomial functions are continuous (Ref: http://www.themathpage.com/aCalc/continuous-function.htm). So, without any maximum or minimum, the given function must be one-one. 

I cant understand this part.why did you do this part.another thing I read any cubic function is onto..is that true???.and is all polynomial function is continious??.sorry for disturbing you..

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Yes. The derivative must be always positive or always negative for a polynomial function to be bijective.

The second part is not needed for this question. Can be used only if the given function is not bijective.

Cubic function is always bijective. Because its derivative is a square term and hence always positive.

All polynomial functions are continuous.

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Thank you sir .

Answer 2

f(x)=x^3 - 3x^2 + 5x - 10
differentiate and check function in strictly increasing or decreasing .

Comments

After that???

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if function is monotonic then it must b one one..
for onto put two extremes of domain(because function  monotonic) and see the range of value coming as output. if set of these values equal to range of Function then that ll be onto.

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please solve it..it is giving complex roots..@arjun sir please give me a detailed solution.I need it.