Question

Find the critical points of : f(x)=|2x+3|

I know I have to find the derivative and set it equal to zero but I am having trouble. I found the derivative as (4x+6)\((sqrt(2 x+3))^2)

Answer 1

notice that |2x+3| is has a V shape and the derivative does not exist at x = -3/2. So -3/2 is the critical number

Answer 2

I think you'd find the problem easier to understand if you'd re-write |2x+3| as a piecewise-defined function, with the left piece defined on (-infinity, -3/3) and the right piece on (-3/2, +infinity). Done this before?

Then you need to think carefully about what "critical point" means in this situation. Hint: look very carefully at what happens to the slope of this function as you move from left to right and cross the vertical line x=-3/2. You might want to graph both the function and its 2-part derivative graph.

Answer 3

f(-3/2) = 0.

so critical point is (-3/2,0)

Answer 4

I found someone else found the critical points as 2 and -2? is that possible?