Are f(x)=ax+cos(x) and g(x)=ax+sin(x) Invertible for a<-1?

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For the functions f(x) = ax + cos(x) and g(x) = ax + sin(x) to be invertible when a < -1, both functions must be monotonic. This is established by demonstrating that their derivatives, f'(x) = a - sin(x) and g'(x) = a - cos(x), do not change sign. Since a < -1 ensures that f'(x) < 0 and g'(x) < 0 for all x, both functions are strictly decreasing and thus invertible. The domains of both functions are all real numbers, and their ranges extend to negative infinity.

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If a<-1 show that f(x)=ax+cosx and g(x)=ax+sinx are invertible functions;(What are their domain of definitions and ranges?)
 
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Can you demonstrate that for $a<-1$ both functions are monotonic, thus invertible?
 
MarkFL said:
Can you demonstrate that for $a<-1$ both functions are monotonic, thus invertible?

How to demonstrate that?
 
What condition must hold in order for a function to be monotonic?
 
MarkFL said:
What condition must hold in order for a function to be monotonic?

For all x<y,f(x)<f(y)?
Or for all x>y,f(x)<f(y)
 
What is true about a function's derivative if it is monotonic?
 
MarkFL said:
What is true about a function's derivative if it is monotonic?

f'(x)<0 or f'(x)>0
 
Good, yes, this is what is required for strict monotonicity. As long as the derivative has no roots of odd multiplicity, then the function is monotonic.

Can you show then that for $a<-1$ that the derivatives of the two functions will never change sign?
 
MarkFL said:
Good, yes, this is what is required for strict monotonicity. As long as the derivative has no roots of odd multiplicity, then the function is monotonic.

Can you show then that for $a<-1$ that the derivatives of the two functions will never change sign?
like using the graph?
 
  • #10
I would do it algebraically. For the first function, we are given

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

Differentiating, we find:

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

Next, I would begin with:

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

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

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