Question

photo attahced.
help please

Answer 1

for the derivative, use the quotient rule.

Answer 2

yep, i know that,...

Answer 3

is that what you got? your numerator seems to have some problem there

Answer 4

numerator is (x^2+4)^2

Answer 5

what?
no I meant the derivative that you have in the picture is wrong.

Answer 6

o.. I see hold on i will cauclate again

Answer 7

f(x)''= (x^2+4)^2(56x)-(56x)(x^2+4)/(x^2+4)^4

Answer 8

what about the first derivative?

Answer 9

the question talks about the first derivative, not the second.

Answer 10

\[
{d\over dx}\left(u\over v\right)=\frac{u'v-uv'}{v^2}
\]

Answer 11

Sorry! I posted the wrong photo
this is it

Answer 12

@electrokid

Answer 13

ok.. part (a) seems wrong.

Answer 14

\[
f'(x)=\frac{(14x)(x^2+4)-(7x^2)(2x)}{(x^2+4)^2}
\]

Answer 15

no. its ok. sorry fr the confusion.

Answer 16

so, for the second derivative..

Answer 17

yes, the second one that I have been sturggling with! part (b)

Answer 18

hold on..
you have a critical number...
there must be an increasing and decreasing region split by it.

Answer 19

take \(x=-1\) what is \(f'(-1)\)?

Answer 20

some negative number.. agree?

Answer 21

also decreasing and increasing problems... too.

Answer 22

your minima is correct.
to have a minima, that means the function should first decrease and then increase, correct?