Closed Interval Method

[stinajeana]

For each pair of function f and an interval [a,b] on which f is continuous, use Closed Interval Method to find the absolute max and absolute min values of f on [a,b]

1. f(x)= 3sqrt(4x^2-4x) [-1,2]

...(the 3 is in the index of the square root!)

So I know that I have to find the critical numbers first and then plug the critical numbers into the original formula along with the -1 and 2 in order to find the absolute max and min. I know it's really simple question.. like I understand how to solve this question and I understand the whole point of solving it but the thing that i'm having trouble finding the derivative. I'm struggling with the derivatives (I know, for you guys it's easy) what I got so far is:

(4x^2-4x)^(1/3)

1/3(4x^2-4x)^(1/3-1)

 

[daon2]

For each pair of function f and an interval [a,b] on which f is continuous, use Closed Interval Method to find the absolute max and absolute min values of f on [a,b]

1. f(x)= 3sqrt(4x^2-4x) [-1,2]

...(the 3 is in the index of the square root!)

So I know that I have to find the critical numbers first and then plug the critical numbers into the original formula along with the -1 and 2 in order to find the absolute max and min. I know it's really simple question.. like I understand how to solve this question and I understand the whole point of solving it but the thing that i'm having trouble finding the derivative. I'm struggling with the derivatives (I know, for you guys it's easy) what I got so far is:

(4x^2-4x)^1/3
1/3(4x^2-4x)^1/3-1

Chain rule: \(\displaystyle \dfrac{d}{dx}(g(x))^{1/3} = \frac{1}{3}(g(x))^{-2/3}\cdot g'(x)\)

 

[stinajeana]

Chain rule: \(\displaystyle \dfrac{d}{dx}(g(x))^{1/3} = \frac{1}{3}(g(x))^{-2/3}\cdot g'(x)\)

so 1/3(4x^2-4x)^(-2/3)*(8x-4)

How would I simplify it in order to find the critical numbers?

 

[daon2]

so 1/3(4x^2-4x)^-2/3*(8x-4)

How would I simplify it in order to find the critical numbers?

Negative exponents tell you to toss things into the denominator. You'll get

\(\displaystyle \dfrac{8x-4}{3(4x^2-4x)^{2/3}}\)

 

[stinajeana]

Negative exponents tell you to toss things into the denominator. You'll get

\(\displaystyle \dfrac{8x-4}{3(4x^2-4x)^{2/3}}\)

which then equals 4(2x-1)/(4x^2-4x)^(2/3)

So one of the critical numbers would be -1/2 ... and then with the denominator would be 4x(x-1)^(2/3)?

 

[mmm4444bot]

So one of the critical numbers would be -1/2

That's close, but not correct. Please check (or show) your work, on that calculation.

and then with the denominator would be 4x(x-1)^2/3?

No, you may not simplify a radical by doing that. :cool:

The critical numbers are where the derivative is zero or is undefined.

Division by zero is undefined, so think about the values for x that would cause that derivative to be undefined.

 

[stinajeana]

That's close, but not correct. Please check (or show) your work, on that calculation.




Not sure what this incomplete question means. :cool:

I meant 1/2!! I accidentally put a negative. So one of the critical numbers would be 1/2 due to the fact that I put 8x-4=0 and solved for x. I'm stuck on how to solve the denominator now... and I know that this is basic stuff that it is very easy and like I said I understand how to solve it and what i'm solving it for but it's the little things like solving for x in the denominator that are throwing me off.

 

[mmm4444bot]

Ah -- I edited my post, after you quoted it, so not sure if you saw that.

Do you realize that values which make the inner function (4x^2-4x) zero will cause the derivative to be undefined?

This is because 0^(2/3) is zero.

 

[stinajeana]

Ah -- I edited my post, after you quoted it, so not sure if you saw that.

Do you realize that values which make the inner function (4x^2-4x) zero will cause the derivative to be undefined?

This is because 0^(2/3) is zero.

So because the denominator is undefined the only critical number would be 1/2?

 

[mmm4444bot]

Nope.

The derivative is a fraction, and you already found where it's zero (when x = 1/2).

Now you want to know where it's undefined.

It's undefined when the denominator is zero.

What values of x make the denominator zero?

 

[mmm4444bot]

\(\displaystyle \displaystyle \frac{8x - 4}{3\sqrt[3]{(4x^2 - 4x)^2}}\)

Explained another way: The derivative above is undefined, when the denominator is zero, because division by zero is undefined.

The denominator will be zero, when the radicand is zero. The radicand is what we call the expression inside the radical sign: (4x^2-4x)^2

In other words, the following expression is undefined because taking the cube root of zero gives back zero.

\(\displaystyle \displaystyle \frac{8x - 4}{3\sqrt[3]{(0)^2}}\)

So, what values of x make that radicand (and, hence, the denominator) zero?

 

[stinajeana]

Nope.

The derivative is a fraction, and you already found where it's zero (when x = 1/2).

Now you want to know where it's undefined.

It's undefined when the denominator is zero.

What values of x make the denominator zero?

0?

 

[daon2]

0?

Instead of guessing, set the denominator equal to zero and solve for x!

 

[stinajeana]

Instead of guessing, set the denominator equal to zero and solve for x!

(4 plus or minus sqrt16)/8 = 1 and 0

 

[mmm4444bot]

(4 plus or minus sqrt16)/8 = 1 and 0

You need to type grouping symbols where needed.

Yes, the denominator in the derivative is zero when x is 0 or 1.

So, the derivative of function f is zero when x=1/2 and it is undefined when x=0 or x=1.

So, you now have all three critical numbers: 0, 1/2, 1

What's next, in this exercise? :cool:

 

[stinajeana]

You need to type grouping symbols where needed.

Yes, the denominator is zero when x is 0 or 1.

So, the derivative of function f is zero when x=1/2 and it is undefined when x=0 or x=1.

So, you now have all three critical numbers: 0, 1/2, 1

What's next, in this exercise? :cool:

Finding absolute max and min... the domain is [-1,2] so because 0,1/2,1 are in the domain I just take the original equation and make a table of values and plug -1, 0, 1/2, 1, and 2 into the original formula. Then i'll get the absolute max and min.

 

[mmm4444bot]

That's a good plan.

 

[stinajeana]

That's a good plan.

would the absolute max be 2 at -1 and 2 and the absolute min be -2 at 1/2?

 

[mmm4444bot]

would the absolute max be 2 at -1 and 2

Yes, but you don't need to report the "at -1 and 2" part. Just say, "The absolute max is 2".

and the absolute min be -2

No -- check (or show) your work, on f(1/2).

 

[stinajeana]

Yes, but you don't need to report the "at -1 and 2" part. Just say, "The absolute max is 2".




No -- check (or show) your work, on f(1/2).

f(1/2)= 3sqrt4(1/2)^2-4(1/2) = -1