Question

If f'(x) = (x-2)(x-3)²(x-4)³ , then f has which of the following relative extrema?
I. A relative maximum at x=2
II. A relative minimum at x=3
III. A relative maximum at x=4

a)I only b)III only c)I and III only d)II and III only e)all

Answer 1

b) III only
See if f' changes signs at each of the points, if they do, then there is an extrema.

Answer 2

Consider taking the second derivative as f'(x) = 0 at all those points.

Answer 3

Second derivative test doesn't always work. Its better to see if first derivative changes signs.

Answer 4

If i plug in 2 , 3 or 4 I'll get 0 so do I try testing values such as 2.5, 3.5 as the x values?

Answer 5

It would be simpler to have done a rough sketch of the original function. It would then be "obvious".

Answer 6

How would you go about doing that? I'm not sure how to sketch a function like this one.

Answer 7

test points in each interval: x < 2, 2 < x < 3, 3 < x < 4, and x > 4... the derivative changing signs would be indicative of maximums or minimums at the critical point between two tested points

it would be a 7th degree polynomial function, i dont know if that'd be much easier

Answer 8

You plot points for (say 0,1,2,3,4) but roughly and draw a smoothish line that passes through them. It doesn't work well if you have a function which is changing really rapidly.

Answer 9

BTW If your 7th degree polynomial has about 5 or 6 terms - don't do it! If it is in a factored form, then it's not so bad.

Answer 10

I have tested the values 1,2,2.5,3,3.5,4,4.5 I plugged them into the f'(x) and got positive, 0, negative, 0, negative, 0 , positive found only maximum at 2 and minimum at 4

Answer 11

A plot is attached.

Answer 12

Thank you so much robtobey, the plot proved my test was correct and there's a maximum at x=2 therefore, the answer should be a)I only ; correct me if I'm wrong :x

Answer 13

Sorry it took so long. A Mathematica presentation is attached.