Question

show that, f(x) = x^3 is decreasing on [-1,1]

Answer 1

Sorry, x^3 is an increasing function for all x!

Reason is because the derivative of x^3 is 3x^2 and 3x^2 is positive when x> 0 and when x <0.

Answer 2

Where did you get that question from?

Answer 3

if it is x^3+(1/x^3)
then

Answer 4

Now, you are asking a new question.

Use what you have learned to determine the answer to your next question.

Answer 5

i first calcualte it s derivative
vat after that

Answer 6

set derivative equal to zero or where it is undefined . That will give you any critical points. Set up your chart..and where f'(x) > 0, the graph is increasing, and where f'(x) < 0, the graph is decreasing.

Answer 7

That's the First Derivative Test.

Answer 8

just give me help how to start it with

Answer 9

OK. Watch.

Answer 10

f(x) = x^3 + 1/x^3

First thing I would do is to rewrite the function as

F9x) = x^3 + x^(-3) becuase it is easier to take the derivative. Note that x^-3 = 1/x^3

So the derivative of x^3 + x^(-3),
f ' (x) = 3x^2 -3x^(-4)

Answer 11

So far, so good?

Answer 12

till here,i have already done

Answer 13

Now, we want to take the derivative and find out where does f ' (x) = 0 and where is f ' (x) undefined.

Answer 14

The best way to solve, may be to rewrite the derivative as

3x^2 - 3/x^4

And that is (3x^6 - 3)/x^4

Now, the fraction is zero when the numerator is zero: 3(x^6 - 1) = 0, so x = 1 and x = -1 is the solution.

The derivative is undefined when x = 0 (becuase the denominator will be zero).

So there are 3 critical points: x = -1, 0, 1.

Answer 15

Now we look at four intervals:

where x < -1
where -1 < x <0
where 0 < x < 1
where x > 1

Take a number in each interval and detrmine if f ' (x) is positive or negative. Example, in the first interval, where x < -1, take x = -2..if you plug in x = -2 into f ' (x) you get a positive number.

Do the same for all 4 intervals.

Where the derivative is positive, the function is increasing. So we know that when x < -1, the function is increasing.

So work out the rest of the problem for each of the three remaining intervals.

Answer 16

makes sense?

Answer 17

okay

Answer 18

fine.

Answer 19

Want to know another way to do it?

Answer 20

@msingh Do you?

Answer 21

yes, if there is another way i will also go for it.

Answer 22

By definition. \(f(x)\) is increasing on an interval \([a,b]\) if \(\forall x_1,x_2\in [a,b]\):\[
x_1<x_2 \implies f(x_1)<f(x_2)
\]
Start with: \[
x_1<x_2
\]Now consider how \((x_1)^2\) is positive as well as \((x_2)^2\), since they are squared.

There are two cases.

Case 1: suppose \((x_1)^2<(x_2)^2\):
\[
(x_1)(x_1)^2<(x_2)(x_1)^2<(x_2)(x_2)^2\implies (x_1)^3<(x_2)^3\implies f(x_1)<f(x_2)
\]
Case 2: suppose \((x_2)^2<(x_1)^2\):
This is only possible if \(x_1\) is negative, and will flip the inequality.
Multiply by \(x_1\) then gives: \[
x_2(x_2)^2>x_1(x_2)^2>x_1(x_1)^2\implies f(x_2)>f(x_1)\implies f(x_1)<f(x_2)
\]
So for all casese:\[
x_1<x_2 \implies f(x_1)<f(x_2)
\]

Answer 23

I think the question should be changed to: show that, f(x) = x^3 is increasing on [-1,1]

Answer 24

It's actually increasing everywhere.

Answer 25

sry bro, qquestion was
pt, f(x) = x^3 +( 1/x^3 ) is decreasing on [ -1, 1 ]

Answer 26

I showed you this an hour ago.