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#### renyikouniao

##### Member

- Jun 1, 2013

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How to demonstrate that?Can you demonstrate that for $a<-1$ both functions are monotonic, thus invertible?

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For all x<y,f(x)<f(y)?What condition must hold in order for a function to be monotonic?

Or for all x>y,f(x)<f(y)

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f'(x)<0 or f'(x)>0What is true about a function's derivative if it is monotonic?

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Can you show then that for $a<-1$ that the derivatives of the two functions will never change sign?

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- Jun 1, 2013

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Can you show then that for $a<-1$ that the derivatives of the two functions will never change sign?

like using the graph?

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\(\displaystyle f(x)=ax+\cos(x)\)

Differentiating, we find:

\(\displaystyle f'(x)=a-\sin(x)\)

Next, I would begin with:

\(\displaystyle -1\le-\sin(x)\le1\)

Can you get $f'(x)$ in the middle, and then use $a<-1$ to show that $f'(x)<0$?