Posted by **Anonymous** on Sunday, December 18, 2011 at 1:06pm.

6x-4cos(3X)Critical Numbers

- Calclus -
**bobpursley**, Sunday, December 18, 2011 at 1:17pm
There is no function here. Is there an equal sign somewhere here? A function? What is your question?

- Calclus -
**Steve**, Sunday, December 18, 2011 at 6:12pm
presumably you meant

y = 6x - 4cos(3x)

y' = 6 + 12sin(3x)

y'' = 36cos(3x)

Now, I'm sure you know all about 1st and second derivatives and how they relate to critical numbers:

y has a max/min where y' = 0 and y'' not zero

y has an inflection point where y'' = 0

y is increasing where y' >. 0

y is concave up when y'' > 0

No go plug and chug

## Answer This Question

## Related Questions

- Math - 2x^3+3x^2-120x to find the critical numbers I took the derivative which ...
- Calculus - How do you find the critical numbers and the end points of this ...
- critical points - (1 pt) Find all critical values for the function f(x) = 6 x^3...
- critical numbers - How would I find the critical numbers of f'(x)=(x-3)^3(x+4)(x...
- Math - The critical numbers of the function: f(t)=7t^(2/3)+t^(5/3) f'(t)=14/3t...
- Calculus - Find the critical numbers of the function g(t)=|5t-7| If you find g'(...
- Math Analysis - what are the two critical points for r=4cos( 4(theta)) and its ...
- help please! - f(x)=x+1/x Find all critical numbers. Find where the function is ...
- Calclus - What is the derivative (tˆ2)ˆ(1/3)+2(tˆ5)ˆ(1/2)
- Critical Numbers - Find the critical numbers of the function on the interval 0...

More Related Questions