Find the critical points of the given functions.

Question

I just need to confirm something

y = (2x − 1)²

find x₁ 2(2x-1) d/dx 2x-1

y' = 2(2x-1) * 2

y' = 4(2x-1)

y'= 8x-4

4= 8x

= 4/8 or 1/2

so x₁ is 1/2

then for y₁

_{y= (2(1/2)- 1)}²

y = (1-1)²

y= (0)²

y₁= 0

My question is can Y₁ Be equal to 0?

Answer 1

The answer to your question is yes, The y coordinate of a point where the x coordinate is one of the critical numbers of a function can be 0. In this particular case the function is a parabola with the vertex located at (1/2, 0). And that is where the slope of the tangent line is zero.

Here is a suggestion when solving for a critical number. When you end up with something like y' = 4(2x-1) then set equal to 0:

4(2x-1) = 0

Consider dividing both sides of the equation by 4 (instead of distributing:

2x-1=0

2x=1

x = 1/2

Here is a Desmos graph for this problem. Use the slider on x₁ to see the tangent line at different points on the parabola. Use the Set link to set x₁ to 1/2.

desmos.com/calculator/dvgrl6z2b6

Answer 2

y=(2x-1)^2= 4x^2-4x+1 = 4(x^2-x+1/4)+1-1=

y= 4(x-1/2)^2+0 in vertex form with vertex= (1/2, 0) = the critical point

it's an upward opening parabola,with minimum y=0

or

y'=8x-4=0

x=1/2

y= (2(1/2)-1)^2=0