f(x)= 3x^2+4x+1

f'(x) = 6x +4

the critical value is x values in which f'(x)=0

then 6x+4=0

==> x=-4/6=-2/3

the the critical vaue for f(x) is x=-2/3

Now f'' = 6 which is positive.

then x=-2/3 is a minimum value for the function.

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now