You know that for some fix value of x. What you need to determine is first the value of x when the function has its minimum value, i.e. when . Plug that value of x into the equation above and you'll get k.
Hi, after a long break I've decided to look at my math hw since I go back to school tomorrow. Well, I don't understand the first problem so hopefully if I can get help with this one I can understand the rest.
1. (noncalculator) The minimum value of f(x) = x^4 - 4x^3 + k is 7. Determine the value of k
I know how to find min and max values, but this is different so all help is appreciated.
Thanks
Oh... completely forgot about that, so if k is a constant then the first derivative is simply 4x^3 -12x^2, and when I set that equal to zero I get x=3 and x=0
I now test to see which value of x gives the min by plugging into the equation of second derivative which is: 12x^2 - 24x
when i plug both of these values into the equation of second derivative i get a value of 0 for when i plug in 0 and a value of 36 when i plug in 3, since 36 is positive it indicates concave up which indicates a min, thus x=3 must be my value for x
I then plug 3 into x for original equation and for k I get -20
does this sound correct??