Thread: relative min

let f be the function given by f(x)=x^4/16-x^3+3x^2/8-2x. The function f has a relative minimum at x=
a. 0
b. 8
c. 11.8
d. 12
e. 15.74

Solve for f'(x) = 0 and use your first derivative test to determine which solution represents a minimum. If you're having trouble, post what you have done and please try to use brackets.

i did the first deriv test to get f'(x)= 1/4x^3 -3x^2 +3/4x-2
but then when i graphed on the calc, there were no zeroes. So does this mean that it's A?
if there are no zeroes, how can i test points to find local max's or min's?

Try zooming out a bit more.

Originally Posted by andrewsx

i did the first deriv test to get f'(x)= 1/4x^3 -3x^2 +3/4x-2
but then when i graphed on the calc, there were no zeroes. So does this mean that it's A?
if there are no zeroes, how can i test points to find local max's or min's?

That the answer is c. is clear from a graph (using a graphics calculator) of y = f(x). Note that the value of y at x = 11.8 is -402.6 ........

I doubt you'd be expected to do this question in a technology free environment.