Calc/Trig

[Kushballo7]

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 + cosx

I think to solve this problem I need to find the derivative. And then
find the minimum or max (I dont know which one) and then use that to
appopriately shift the graph up or down so it doesnt cross the x-axis
[which if it did would yield ciritcal numbers]...I got lost in there
somewhere...help please!

 

[stapel]

Since the critical numbers are the zeroes of the derivative, yes, you need to find the derivative. Since the question does not ask you to find the max/min values, there would be no reason to attempt to find such.

Instead, since it asks for the value(s) of "a" such that there are no critical numbers, it might be better to find the value(s) of "a" such that the derivative is never equal to zero.

Eliz.

 

[Kushballo7]

Since the critical numbers are the zeroes of the derivative, yes, you need to find the derivative. Since the question does not ask you to find the max/min values, there would be no reason to attempt to find such.

Instead, since it asks for the value(s) of "a" such that there are no critical numbers, it might be better to find the value(s) of "a" such that the derivative is never equal to zero.

Eliz.

How would you do that!?

 

[pka]

The derivative is \(\displaystyle f'(x) = - 2(a^2 + a + 6)\sin (2x) + (a - 2) - \sin (x)\).
Now you need to find values of a for which f’(x) has no zeros.

 

[stapel]

...you need to find the derivative...[and] find the value(s) of "a" such that the derivative is never equal to zero.

How would you do that!?

Do which? Find the derivative? Set it equal to zero?

Please clarify where you are having difficulty. Thank you.

Eliz.

 

[Kushballo7]

Oh yeah, that probably would be helpful. I can find the derivative. I'm having trouble finding how it would have no critical numbers.

 

[stapel]

I'm not ignoring you. But the actual calculations for this thing appear to be a royal b!tch....

Eliz.

 

[Kushballo7]

So I noticed. I'm liking calc less and less everyday...

 

[Kushballo7]

MY CALC TEACHER MADE A TYPO. THE REAL EQUATION IS...

f(x) = (a^2 + a + 6)cos(2x) + (a-2)x + cos1

I'm thinking this might make things easier.

 

[pka]

Easy is relative: \(\displaystyle f'(x) = - 2(a^2 + a + 6)\sin (2x) + (a - 2)\). I have a suggestion. Graph f’(x), assigning different values to a.
I know so little about graphing calculators. There ought be a way to assign values to to a parameter a.
You should be able to convince yourself that there are no values of a for which f’(x) is never zero.

In the graph below, I have graph the function, and it first and second derivatives for a=−2. For any a the graphs look much alike.