Question

It is given that f(x) = 1/x^3 − x^3, for x > 0. Show that f is a decreasing function.

Answer 1

hint :
f ' (x) < 0
can u determine f ' (x), first ?

Answer 2

with f ' (x) is the first derivative of f(x)

Answer 3

@accio.firebolt

Answer 4

(-3x^-4) - 3x^2

Answer 5

@RadEn

Answer 6

yup, that's right
or it can modif be :
-3/x^4 - 3x^2, right ?

Answer 7

or
-3(1/x^4 + x^2), right ?

Answer 8

@RadEn so because it's negative, then it's less than 0 so it's a decreasing function?

Answer 9

yes, because the terms in bracket absolutly always be a real positive numbers)
-3 multiplied by a positive number = negative number, in other words always < 0

" that's why f is a decreasing function "

Answer 10

@RadEn thanks so much! is okay if i ask you another question?

Answer 11

if i can i wll :)

Answer 12

dy/dx = 2(3x+4)^1.5 -6x-8
It is now given that the stationary point on the curve has coordinates (-1.5). Find the equation of the curve.

So i tried to find y and i got 4/15 ( 3x +4)^2.5 -3x^2 - 8x + c
Then when i tried to put the coordinates in. the c i got was different from the real answer.

Answer 13

please answer my question

Answer 14

@RadEn

Answer 15

sorry the conection is lagging :(
hmm.. ur integration is okay
yeah, for x=-1.5 gives f(x) becomes not real

Answer 16

are u sure the point given is x = -1.5 ?
it does not exist for the curve of f(x)

Answer 17

the x is -1 and y is 5 for the stationary point

Answer 18

\[{dy \over dx} = 2(3x + 4)^{3 \over 2} -6x -8\]
\[\int \frac{dy}{dx}dx = 2\int \left((3x + 4)^{3 \over 2}-3x -4\right)dx\]
\[y=\frac{4}{15}\left(3x + 4\right)^2\sqrt{3x + 4} - 3x^2 - 8x + c\]

Given that x = -1 and y = 5,
\[\Large 5 = \frac{4}{15}\left(1\right)^{\frac{5}{2}}-3(-1)^2+8+c\]
\[5 = \frac{4}{15} + 5 + c\]
\[c = -\frac{4}{15}\]

Answer 19

@meepi i also got -4/15 for c but the answer is apparently 5

Answer 20

if it's 5, then substituting in -1 gives 154/15 :/
It can't be 5, I'd assume the book is wrong or that you could have interpreted the problem incorrectly

Answer 21

okay thanks anyway :)