Use both the first and second derivative tests to show that 3x^2-6x+1 has a relative minimum at x=1.

Question

Use both the first and second derivative tests to show that 3x^2-6x+1  has a relative minimum at x=1.

freckles · Answer

have you found the derivative of the given expression?

freckles · Answer

first derivative =0 will give us the possible mins and maxs

freckles · Answer

or at least where they occur of course

anonymous · Answer

i derived it and got 6x-6 for the first and 6 for the second derivavtive is thatright?

freckles · Answer

that sounds great.

freckles · Answer

6x-6=0 gives us?

anonymous · Answer

yeah so does that mean 1 is a critical point?

freckles · Answer

yes

freckles · Answer

since since f''>0 then that implies with have what kinda extrema at x=1?

freckles · Answer

Remember I told you earlier what f''>0 implies

anonymous · Answer

relative minimum?

freckles · Answer

yes we have a rel min at x=1

freckles · Answer

$$f(x)=3x^2-6x+1 \ f'(x)=6x-6 \ f'=0 \text{ when } 6x-6=0 \text{ this gives } x=1 \ f''(x)=6>0 \text{ so this implies we have rel min at x=1 }$$

anonymous · Answer

omg that's what I have on my paper! so it is just like the other problem?  nothing els?!

freckles · Answer

it was very similar to your previous problem

anonymous · Answer

cool! thanks!

anonymous · Answer

wait what do I do when I plug in the criticsl point and it equals zero??

freckles · Answer

when f''=0?

anonymous · Answer

oh no never mind lol I plugged it in wrong! but what are stationary points?

freckles · Answer

i think stationary point is another way to say critical point

freckles · Answer

also if x=a is a critical number and f''(a)=0, then this test is inclusive for relative max and min. It just tells us that the function is neither concave up or down at x=a.