# Math Help - critical number?

1. ## critical number?

determine the value of a for which the function f has no critical number:

f(x)= (a2+a-6)cos2x+(a-2)x+cos1

2. ## Re: critical number?

What do you find when you equate the derivative to zero? Is the last term actually cos(1)?

3. ## Re: critical number?

yes the last term is actually cos(1) and how do you take the derivative and set it equal to zero? is a just a constant?

4. ## Re: critical number?

Yes, you would treat $a$ as a constant.

5. ## Re: critical number?

i got the derivative being

-2a^2sin2x-2asin2x+12sin2x+a-2 and then i dont know how to take the derivative of cos1 and i dont know how to set this derivative equal to 0 with the constants in it?

6. ## Re: critical number?

cos(1) is a constant, so its derivative is zero. I would write the derivative as:

$f'(x)=-2(a^2+a-6)\sin(2x)+(a-2)=-2(a+3)(a-2)\sin(2x)+(a-2)=$

$(a-2)(-2(a+3)\sin(2x)+1)=0$

What condition on $a$ causes the second factor to have no root?

7. ## Re: critical number?

i dont see anything where you said i would write the derivative as:
Originally Posted by MarkFL2
cos(1) is a constant, so its derivative is zero. I would write the derivative as:

8. ## Re: critical number?

I accidentally hit Post Quick Reply rather than Go Advanced...I have edited my post.

9. ## Re: critical number?

i dont know what you mean about the root and i still dont understand how to set it equal to zero and solve for whatever like that?

10. ## Re: critical number?

A root is a value that causes an expression to be zero. Anyway, consider:

$-1\le\sin(2x)\le1$

Now, equate the second factor to zero, solve for $\sin(2x)$ and substitute into the above inequality. This will tell the the values of $a$ for which there are critical numbers.

11. ## Re: critical number?

so i got a=2 and then the second part (a+3)sin(2x)=1/2 and i dont know how to simplify it any further

12. ## Re: critical number?

When a=2, your function is constant, so there will be no critical values.

As for the second part, solve for sin(2x), then as an alternate to what I suggested before, set the magnitude of this to greater than 1. You will have 2 cases to consider:

i) $\frac{1}{2(a+3)}>1$

ii) $\frac{1}{2(a+3)}<-1$

13. ## Re: critical number?

i dont understand

14. ## Re: critical number?

We found:

$\sin(2x)=\frac{1}{2(a+3)}$

We therefore know that if $\left|\frac{1}{2(a+3)} \right|>1$ then there is no solution, that is, there is no real value of $a$ which will give us a critical value.

15. ## Re: critical number?

well how do you solve for a?

Page 1 of 2 12 Last