Question

the derivative of f is given as 2(x-1)(2x-1)^2(5x-2). f has a relative maximum at
a. x=1 only
b. x=2/5 only
c. x=1/7 only
d. x=-1/2 only
e. x=1, x=-1/2

Answer 1

Extrema of f(x) is/are the solution(s) of f'(x)=0

Answer 2

so do i expand the given derivative? or am i able to solve with the function as is?

Answer 3

If f''(x) is positive, x is a minimum. If f''(x) is negative, x is a maximum.

Answer 4

If f''(x) is zero, check your calculation xD

Answer 5

@laniraegreen just solve the equation f'(x)=0 :) (P.S. f'(x) is given)

Answer 6

but to solve for an x value, do i need to fully multiply the given f'(x)?

Answer 7

no you don't, f'(x) is already factorized to facilitate you solving it :)

Answer 8

If something=0 and "something" is factorized, you set each factor to zero.

Answer 9

That means x-1=0 or 2x-1=0 or 5x-2=0

Answer 10

okay! i understand that now.

Answer 11

remember to plug the solutions to f''(x) to see if it's a minimum or a maximum :)

Answer 12

so i have my three solutions, but do i differentiate each factor to get f''?

Answer 13

i know how to factor, i just don't know the next step

Answer 14

differentiate* haha sorry

Answer 15

You either:
use the product rule; or:
expand it then differentiate

Answer 16

get it?

Answer 17

i'm just substituting the solutions into f''.....

Answer 18

I have to go now, so hopefully someone will help you :)
@ganeshie8 @Zarkon

Answer 19

okay, thank you so much for all your help though! :)

Answer 20

once i get my values, then what? is the negative value the maximum?

Answer 21

since f'(x) is in factored form, you may try and visualize what intervals f'(x) is increasing/decreasing and figure out if f"(x) will be +ve/-ve/0

Answer 22

2(x-1)(2x-1)^2(5x-2)

degree = 4, use the precalculus concepts. wat can you tell about end-behavior ?

Answer 23

that both ends will extend into either positive or negative infinity?

Answer 24

yup ! since the coefficient is +ve, both ends go UP

Answer 25

so theres one maximum?