Question

Calculus!!!
Please helpp!!!!
Find the absolute minimum of the given function on the specified interval. Give your answer as simplified fractions.

f(x)=(1/3x^3)-36x+27 ; 3≤x≤6

Absolute minimum value=____ when x=_____

Answer 1

1) Find your "critical numbers"
To do so, find which values of x will make the derivative = 0.
So do the derivative and set it equal to zero to find your critical value(s).
Your critical numbers are also the end points of your interval (3 and 6 in this case)

2) Now that you have your critical numbers, plug them into the original equation and find out which critical number gives you the lowest value of y.
The absolute min value is the value of y that is smallest of the critical numbers x possible.
Likewise, if you needed to find the absolute MAX, you use your critical numbers, plug them into the original equation, and solve for the LARGEST of the values.

For example:
\[y = x^2 + 2x\]
it's derivative is
\[y ' = 2x + 2\]
set equal to zero
\[2x + 2 = 0 ==> x = -1\]
-1 is a critical number

if my interval is -3<x<0, -3 and 0 are also critical numbers.

you have -3, -1, and 0 as critical numbers, so plug them into the original equation.
for -3
\[y = (-3)^2 + 2(-3) = 3\]
for -1
\[y = (-1)^2 + 2(-1) = -1\]
for 0
\[y = (0)^2 + 2(0) = 0\]

so you have 3 points:
(-3,3), (-1,-1), and (0,0)
Whichever has the lowest y value, that's where the min is.
so the absolute min value is -1 when x = -1

Answer 2

Ok, I understand that. But when i plug it in it says wrong answer..

Answer 3

This is my forth attemp and still the answer for the absolute values are way too high like -205 and the other part like when x= its 5 in this case.

Answer 4

I meant 6..

Answer 5

@JakeV8 Do you have an idea of how to solve this?
Why the answerr is Absolute minimum values=-334/3 when x=5.?
Please help me!!

Answer 6

(working on it...)

Answer 7

is the first term 1/ (3x^3)

or is it (1/3)x^3

Answer 8

\[\frac{ 1 }{ 3x ^{3} }, or, \frac{ x ^{3} }{ 3 }\]

Answer 9

(1/3)x^3

Answer 10

ok, so the derivative is x^2-36, right?

Set derivative = 0... x^2 - 36 = 0

Factor: (x+6)(x-6) = 0

So x = -6 and x = 6 are points at which slope is 0, meaning they are either max or min points along f(x).

However, the problem states that you need to look on the interval from 3 to 6, so you can throw out the x = -6 possibility.

Answer 11

Now you know that a max or min point occurs at x=6.

Next, plug x = 6 back into f(x), to get f(6) = (1/3)(6^3) - 36(6) +27 =
= 72 - 216 + 27 = -117

So the point (6, -117) should be the minimum (or the max) within that region of x between 3 and 6.

To check, you could try x=3, x = 4, x = 5 and find f(x) at each.

But (6,-117) should be correct.

Answer 12

Good explanation I loved it>>

Answer 13

well, hold on. maybe that's a local min but not the "absolute min"?

Answer 14

you think the min occurs at x = 5?

Answer 15

But why the answer is F(6)=334/3 when x= 5

Answer 16

huh?

Answer 17

Yes< I been trying for so long thinking that I have the right answer.. and it came out like that

Answer 18

for x = 5, I get f(5) = (1/3)5^3 - 36(5)+27 = -111.333333 which could be written as -334/3

Answer 19

Yes is minus!! -334/3

Answer 20

it's approx -111, so it's negative but not as far down as f(6)=-117, or (6,-117)

Answer 21

ok, so that is a value at x = 5, but at x = 6, the function is more negative.

x=6 appears to me to be the x-coordinate of the min f(x) point in that region, so (6,-117) should be it.

Answer 22

So hold on you plug in numner 5, and why you got -111.33333. so how we convert it into the answer?

Answer 23

like -334/3?

Answer 24

why are you thinking we need to use x = 5? That part is confusing me...

If I were responding, I think my answer would be:

Absolute minimum value=____ when x=_____

Absolute minimum value= -117 when x= 6

Answer 25

I just the answer the question and the right answer was
Absolute minumum value = -334/3 when x=6.... Sorry no 5...

Answer 26

sheesh.

Answer 27

jejeeje!!

Answer 28

My question is now.. how can convert this fraction.

Answer 29

I am trying another one with different numbers now and my calculator is giving me -320 2/3

Answer 30

I plugged in this case will be number 7. into the equation. f(x)=(1/3x^3)-64x+13 ; 5≤x≤7

Answer 31

Is there any chance on earth that you mis-typed part of the question, and that the correct region for x should have been 3 <= x <= 5 ?

Answer 32

(my question just now was in reference to the original question we were working, not the new one)

Answer 33

No, I made sure that the qauestion was correct. before putting in.

Answer 34

I doubled check first..

Answer 35

It was my mistake, when I wrote the answer, i typed 5 instead of 6..

Answer 36

This is getting nuts.

Here's the situation they are giving you...

In general, given f(x), you can find max and min points across the entire domain of x value inputs by taking the derivative and setting equal to 0. Those are "flat" spots on the curve, so they are either local max or local min points.

However, on a curve like an x^3 curve, it gets infinitely big positive for positive x and infinitely negative for negative x, but in the middle, there is a little wiggle.

These problems are purposefully giving you a region of x values that includes part of that wiggle. Also, the region of allowed x values may NOT include the x value that would make the derivative equal 0. That means that although the overall function does have a true min or max point somewhere, it doesn't sit inside the region of interest. Instead, you may end up needing to choose the "most negative point in the region" as the absolute min in that region.