Question

Consider the function y = 3x5 – 25x3 + 60x + 1. Use the first or second derivative test to test the critical points. How many relative maxima did you find?

Answer 1

so what did you get for

f'(x)

and f"(x)

Answer 2

15x4-75x2+60=f'
60x3-150x=f"

Answer 3

i think i found the critical points already. but im confused on how to test it to find the maxima.

Answer 4

critical points: –2, –1, 1, 2...??? i think.

Answer 5

ye, thats correct, so now test them in the 2nd derivative by substituting

if f"(a) > 0 you have a minimum

f"(a) = 0 a horizontal point of inflexion

f"(a) < 0 you have a maximum...

so text the solutions and you should be able to find the maxima...

hope it helps

Answer 6

so tests the critical pts in the f'' and follow those rules...?

Answer 7

oops a is a solution to the 1st derivative

Answer 8

yes... for example

f"(-1) = 60(-1)^3 - 150(-1) = 90 so a minimum...

Answer 9

that doesnt make sense. sorry, youre good at guiding me, but im having trouble embedding what youre saying with what ive learned. doest a min have to have a negative numbers?

Answer 10

YES!!! i solved it! it has 2 maximum!!! right?

Answer 11

no a minimum can be any where on the graph because there are relative to each other..

here is a graph of your curve

Answer 12

thats correct... at x = -2 and x = -1

Answer 13

does the original equation have a min at f' when x=1

Answer 14

i mean max

Answer 15

as you can see from the graphs it has mins at x = -1 and x = 2

you can solve the 2nd derivative for possible points of inflexion, and check by testing either side of the points for a change in concavity

Answer 16

oopps yes... a max at x = 1.... not -1

Answer 17

whats the difference between a local and relative min/max

Answer 18

nevermind! i just checked my answers and i got them all right! thank you!