Question

Find the absolute maximum value and the absolute minimum value, if any, of the function

g(x)=x^2+4x^(2/3) on [0,1]

Answer 1

I got to the first derivative, which I believe is:
g'(x)= 2x + (8/3)x^(-1/3)

How do I get the min and maximum from here??

Answer 2

I'm stuck at the part of trying to solve for x when setting the first derivative equal to zero

Answer 3

it might be easier to take the second derivative, and figure out the maximum/minimum points that way

Answer 4

If you have a continuous function on a closed interval, absolute minimums will be either at

a) A critical point (g'(x)=0)
b) At an endpoint of the interval (x=0 or x=1 in this case).

So identify the critical points and calculate the function value at the endpoints and critical points. The lowest of these is the absolute minimum, highest is the absolute maximum.

Answer 5

that way you only have one x term when solving for the critical points, and it is much easier

Answer 6

ok let me try the second derivative.

I have a question. How do I know if I should only use the first derivative, or keep going and do a second derivative test to get the answer?

Answer 7

@funinabox Are you sure that works? f''(x)=0 doesn't give you a critical point.

Answer 8

@beginnersmind no it wont, however, f'(x) = 0 is a critical point, and for any f''(x) > 0 or f''(x) < 0 you will have a critical point (as long as f'(x) = 0 is also true at that point).

it's more or less a work-around. I'm only suggesting it because i didn't see an easy way besides using the quadratic formula to solve for the critical points lol

Answer 9

I don't see a way around solving 2x + (8/3)x^(-1/3)=0
substituting u=x^1/3 gives 2u^3+(8/3)/u=0, which doesn't look too bad.

Answer 10

man this one is getting hairy with all the fractions.. so second derivative is:
(-8/9)x^(-4/3)

Answer 11

now I set that to zero, and find x to see if its within the interval endpoints?

Answer 12

you should also be able to solve for f''(x) = 0, and use revert back to the first derivative test after figuring out what points are actually max/min points, by testing around the points.

this should work cause the second deriv will give you the rate of change of the rate of change, thus after reverting back to the first deriv you will have the same thing as if you started out solving for f'(x) = 0.

however, i'm not sure if the domain of the 2nd derivative does not include all numbers inside it's interval (i.e. there is some number that does not exist)