Question

Calculus:
question below.

Answer 1

where's the question?

Answer 2

At last! I already opened this..
sorry i'm just waiting for OS to load
here I attached my question..

Answer 3

@zpupster can you help me?
@agent0smith

Answer 4

i have been trying but i do not think i quite have it . i think it is asking for proof of the product rule.

Answer 5

i don't think so..?
btw, i just multiplied both of th function, since they are positive I assume that their products are also positive so it's above x axis too.. now i just need to know whether it will increase..
thanks 4 ur effort!

Answer 6

f(x) = -e^(-x) and g(x) = x are both increasing but their product -x*e^(-x) is decreasing.

Answer 7

f(x) = -e^(-x) increasing??

Answer 8

no that does not work

Answer 9

but it says on the question when f(x)> 0 and g(x)>0 which means they are both above x-axis.. f(x) = -e^(-x) is below x-axis..

Answer 10

Yeah i think it might just be looking for something using the product rule... it seems pretty easy though:

(fg)' must be positive, for fg to be increasing.

\[\Large (fg)' = f'g + fg'\]
We're told f and g are both increasing, so f' and g' must be positive.

We know f and g are both positive,

thus (fg)' is positive (since f', g, f, g' are all positive)

Answer 11

@Data_LG2 I'm assuming you understand that an increasing function means that it's derivative must be positive?

Answer 12

yes i know that:)

there's an explanation in the book,
let me type it down..

Answer 13

LEt y=f(x) and u=g(x).
Let x1 and x2 be any two values in the intervel [a, b] so that x1<x2.
Since x1<x2, both functions are increasing:
f(x2)>f(x1)
g(x2)>g(x1)
yu=f(x)g(x)
f(x1)g(x2)>f(x1)g(x2)
therefore yu or fg is strictly increasing..

Answer 14

Hmm... i greatly prefer my explanation :P

Answer 15

i know right;)

Answer 16

I mean their's makes sense... but their's is just algebraic, considering it's a calculus class,and assuming you've learned the product rule, i like a calculus based proof.

Answer 17

yeah, i learned product rule....
that's why post the question here for better understanding of this explanation in the book which i'm not satisfied:/

Answer 18

Theirs basically says that, since f(x2) is bigger than f(x1), and the same goes for g(x2) and g(x1), then the product f(x2)*g(x2) must also be bigger than f(x1)*g(x1) - since f(x2)*g(x2) is a product of two bigger numbers.

Answer 19

but what if f(x)<0 and g(x)<0? is the product fg is still be increasing, decreasing or neither???

Answer 20

go back to this\[\Large (fg)' = f'g + fg'\]
if f and g are both negative, but f' and g' are both positive, what will (fg)' be?

Answer 21

increasing??

Answer 22

You know f' and g' are both pos. and f and g are both neg.

Use that to work out whether each term is pos. or neg. in this: \[\Large (fg)' = f'g + fg'\]

Answer 23

oh it's f ' and g', sorry i did not see the primes..

so they will be negative which means decreasing?

Answer 24

Yep. both f'g and fg' will be negative, so their sum must be negative.

Answer 25

ok.... thanks a lot!!(^_^)