Question

What is the minimum 3x^2+2^6 x^-3

Answer 1

I believe it is 20 when x=2.
(2, 20). Try graphing it if you are allowed to.

Answer 2

How did you do that?

Answer 3

fisrt, can you find the derivative of 3x^2+2^6 x^-3 ?

Answer 4

6x+12^5-3x^-4

Answer 5

I found the derivative and checked it at critical points. When the derivative=0 or when it doesn't exist we have a critical point.

Answer 6

12^5 (part) ? how you got that, @psam24 ?

Answer 7

That critical point can then be used to determine if the point is a maximum or minimum. What we can do is set up a sort of number line known as a 'sign chart' in which we put critical value, and figure out if the points to the left and right are positive and negative.

Answer 8

@RadEn thats the part im confused about what it the derivative of that, just 0?

Answer 9

To find the derivative of 3x^2+2^6x^-3. First use the power rule on 3x^2. Then the quotient rule on 2^6x^-3.

Answer 10

see the original function, let it is f(x) = 3x^2+2^6 x^-3
notice that 2^6 = 64, so f(x) = 3x^2+2^6 x^-3 = 3x^2+64 x^-3

therefore, f '(x)= 6x + 6(-3)x^-4 = 6x - 192x^-4

Answer 11

Now that you have that derivative from Rad, set f'(x)=0 and see what you get.

Answer 12

opss, i mean
f '(x)= 6x + 64(-3)x^-4 = 6x - 192x^-4

Answer 13

@RadEn ok lol i knew something was wrong then what?

Answer 14

Please look at my above statement.

Answer 15

what do you mean from your above statement, can i see it in action?

Answer 16

Here...RadEn helped you get the derivative.
f '(x)= 6x + 64(-3)x^-4 = 6x - 192x^-4
When f'(x)=0 or it doesn't exist, we can get something called a critical point.
Critical points can help us figure out maximum and minimums.
Set the derivative=0 and solve for x.

Answer 17

wow ok that looks a bit tough

Answer 18

Not particularly. 6x-192/x^4=0, start by telling me where this derivative cannot exist. (Hint, can you ever divide by 0?)

Answer 19

how did you get to that next step of 6x-192/x^4=0

Answer 20

Are you in calculus?

Answer 21

We have an equation you gave us. f(x)=3x^2+2^6*x^-3
We have taken the derivative. f'(x)= 6x-192*x^-4
Now we set the derivative equal to 0. f'(x)=0, thus 6x-192*x^-4=0
x^-4 is the same as 1/x^4 so I can also say 6x-192/x^4=0

Answer 22

ok i see

Answer 23

then from there?

Answer 24

First tell me what you get for x so that the derivative is nonexistant.
Then tell me what you get for x when 6x-192/x^4=0

Answer 25

0

Answer 26

Good! Now what do you get for x when 6x-192/x^4=0? Try factoring out an x.

Answer 27

..

Answer 28

f ' (x) = 0
6x - 192/x^4 = 0
6x = 192x^4
times x^4 on both sides, get
6x^5 = 192
divided by 6 both sides get
x^5 = 192/6
x^5 = 32
setting 32 = 2^5, so
x^5 = 2^5
the power both sides is same, therefore the bases must be same too, get
x = 2

subtitute x = 2 into f(x) = 3x^2+64 x^-3 = 3x^2 + 64/ x^3
so, the value of f(2) = 3(2)^2 + 64/(2)^3 = 3(4) + 64/8 =12 + 8 = 20.
that's the minimum value of f(x)