Question

To sirm

Answer 1

@sirm3d

Answer 2

\[f'(x)=2x^{4/5}(x-4)+(4/5)x^{-1/5}(x-4)^2\]\[\quad = (2/5)x^{-1/5}(x-4)[5x+2x-8)\]\[\quad = (2/5)x^{-1/5}(x-5)(7x-8)\]

Answer 3

three critical numbers x = 0 (vertical tangent line) and x = 5, 8/7 (horizontal tangent lines)

Answer 4

isnt't critical number is when y=0, x=0 and x=4..

Answer 5

those are roots. critical numbers are roots of the derivative, NOT roots of the given function

Answer 6

ahh.. and how do u know u need to take out 2/5 ? I know because it gives u whole number, but i have trouble deciding which number to take out in order to get a whole number

Answer 7

you don't have to take out any constant. your objective is to find the roots by setting the derivative to zero.

Answer 8

but u took out 2/5, if u dnt take it out, you have a bunch of fractions and cant factor things out to get critical numbers

Answer 9

not really. i can have \[x^{-1/5}(x-4)[2x+(4/5)(x-4)]\]
setting each factor to zero, \[x=0\]\[x-4=0\]\[2x+(4/5)(x-4)=0\]
the last equation yields x=8/7

Answer 10

oh ok thx! and can u refresh my mind on riemann sum as well?I think i forgot abt it already ..

Answer 11

my notes just look foriegn to me..

Answer 12

okay. what number would that be?

Answer 13

#4 and #10

Answer 14

I actually did #4 on the test , but now just completely forgot....

Answer 15

(+1)+(-1)+(+1)+(-1)+...+? (last term)

Answer 16

i=0 means start from 0 , so (-1)^0 = 1 ,
and continue i=1 (-1)^1 = -1

i=100 is (-1)^100 = 1

so answer would be A, 1 ?

Answer 17

since the last term didnt get cancelled out

Answer 18

yup

Answer 19

ok ty :D and #10 plz :)

Answer 20

C. constants can be moved before a summation.

Answer 21

why is E not true?

Answer 22

i thought the rieman sum of constant is always constant itself?

Answer 23

the right side has one term more than the left side.

Answer 24

what's that mean..? i dont really get those i=0 and i=1 ..

Answer 25

r u still here @@