Question

Locate all critical points of

6x^2-2x+1

can you please explain how as well? plz and thank you

Answer 1

have you done any calculus?

Answer 2

yes, I recently got to this point in time of class, and I forgot how to do it, and Im a bad note taker :/

Answer 3

ok
you first find the derivative of the function which gives your the slope of tangent
at any point on the graph. At a turning point this slope = 0.
f(x) = 6x^2 - 2x + 1
can you find f'(x) ?

Answer 4

f'(x)=12x-2

Answer 5

right
so 12x = 2
x = 1/6
other coordinate y = 6(1/6) - 2(1/6) + 1 = 5/3

so theres a turning point at(1/6, 5/3)
do they want to know its nature ( maximum, minimum os point of inflection)?

Answer 6

they dont want to, right now, but possibly in the future, if you can explan thatll be awesome

Answer 7

ok there's 2 ways
either find the slope before and after the critical point
or check sign of the second derivative