Find a for which function f has no critical number

[issacnewton]

Homework Statement

Determine the values of the number $a$ for which the function $f$ has no critical number. $$f(x) = (a^2+a-6)\cos{2x} + (a-2)x + \cos{1}$$

Homework Equations

Concept of critical point

The Attempt at a Solution

Now the derivative of this function will be $f'(x) = -2(a^2+a-6)\sin{2x} + (a-2)$. Now the amplitude of the sine function is $2 |a^2+a-6|$. The maximum and the minimum values of the first term are $2 |a^2+a-6|$ and $-2 |a^2+a-6|$. So $f'(x)$ will never be zero if $|a-2| > 2 |a^2+a-6|$. And if $f'(x)$ is never zero, then the function $f$ will have no critical number. $f'(x)$ exists for all real numbers, so we don't have to worry about the points at which $f'(x)$ does not exist. So function $f$ will have no critical number if $|a-2| > 2 |a^2+a-6|$. Solving this inequality, we get $-\frac{7}{2} < a < -\frac{5}{2}$. Is this correct ?

 

[haruspex]

Homework Statement

Determine the values of the number $a$ for which the function $f$ has no critical number. $$f(x) = (a^2+a-6)\cos{2x} + (a-2)x + \cos{1}$$

Homework Equations

Concept of critical point

The Attempt at a Solution

Now the derivative of this function will be $f'(x) = -2(a^2+a-6)\sin{2x} + (a-2)$. Now the amplitude of the sine function is $2 |a^2+a-6|$. The maximum and the minimum values of the first term are $2 |a^2+a-6|$ and $-2 |a^2+a-6|$. So $f'(x)$ will never be zero if $|a-2| > 2 |a^2+a-6|$. And if $f'(x)$ is never zero, then the function $f$ will have no critical number. $f'(x)$ exists for all real numbers, so we don't have to worry about the points at which $f'(x)$ does not exist. So function $f$ will have no critical number if $|a-2| > 2 |a^2+a-6|$. Solving this inequality, we get $-\frac{7}{2} < a < -\frac{5}{2}$. Is this correct ?

Looks right. Not sure how you did the last step, but substituting a=c+2 makes it easy.

 

[issacnewton]

haruspex, I considered cases here. Since $(a^2+a-6) = (a-2)(a+3)$, I considered cases where $a < -3$, and $-3 < a < 2$ and $a>2$. Whenever there are abslute values, taking different cases help. But yes, with your substitution, things become little easier.

 

[haruspex]

with your substitution, things become little easier.

Yes, it avoids the need to break into cases. |c|>2|c²+5c|, so 1>2|c+5|.

 

[issacnewton]

Makes sense, haruspex