Question

For the following graphs, state the intervals of x for which the functions are increasing and those for which they are decreasing. Sketch the graphs- (iv) y= 3x^2 - (1/x^2), (x does not equal 0).

Answer 1

... is this calc or algebra?

Answer 2

Calculus again.

Answer 3

oh ok. so take the derivatives. if first derivative > 0 , increasing, <0 decreasing, =0 it's flat / not changing.

Answer 4

That's my problem. the derivative I've got here is x^4+(1/3)

Answer 5

x^4 + (1/3) > 0. How do you solve that?

Answer 6

In these questions, you don't use "i".

Answer 7

well i can't tell ... use the equation editor please

Answer 8

\[\frac{ d }{ dx } x^4 + \frac{ 1 }{ 3 }\]

Answer 9

no i mean the original equation. sorry.

Answer 10

Oops, wait that's the wrong derivative sorry. That was simplified when I made the derivative less than zero.
\[y= 3x^2 - \frac{ 1 }{ x^2 }\]

Answer 11

\[\frac{ d }{ dx } \frac{ 6x^4 + 2 }{ x^3 }\]

Answer 12

huh. oh sorry yes?

Answer 13

by the way the derivative is \[12x-\frac{ 4 }{ x^3 }\]

Answer 14

so personally i don't like that form so let's make it into a combined form ex. a/b where a and b are expressions

Answer 15

Wait. @RONNCC I combined the equation before differentiating?

Answer 16

Are you meant to differentiate then combine?

Answer 17

look at image
if x>0 f increase
if x<0 f dec

Answer 18

f increases when x<0, that's what the answers say.

Answer 19

I agree with your diagram but why does f increase when x<0. That's what the answers say.

Answer 20

well the function is increasing whenever its derivative is > 0, and decreasing whenever its derivative is < 0

Answer 21

The answers say decreasing when x< 0 increasing when x < 0.

Answer 22

if f(x) = 3x^2 - 1/x^2 then
f'(x) =

\[\frac{ 2 }{ x^{3} } + 6x\]

or combined

\[\frac{ 6x^{4} + 2 }{ x^{3} }\]

Answer 23

we know the numerator is always positive so we must look at the denominator. whenever the denominator is positive we have +/+ which is positive

so if x > 0, the derivative is positive

if the denominator is negative we have +/- which is negative

so if x < 0 the derivative is negative

Answer 24

"The answers say decreasing when x< 0 increasing when x < 0"

what does that mean? it cant be decreasing when x < 0 and increasing when x < 0. that is a typo

Answer 25

That's what it says.

Answer 26

it contradicted itself then

Answer 27

Is it meant to be increasing when x>0?

Answer 28

probably lol

Answer 29

thats like saying a number is positive, but also negative

Answer 30

I think it means the derivative is negative when x<0 and the derivative is positive when x<0

Answer 31

i think it means

derivative is negative if x < 0
derivative is positive if x > 0

because that's what the first derivative shows

Answer 32

ah okay