Find the absolute maximum and absolute minimum values of on the given interval : f(t) = t(4 - t^2), [-1, 2]

Question

Find the absolute maximum and absolute minimum values of on the given interval : f(t) = t(4 - t²), [-1, 2]

 

I know I first need to find the y values for the endpoints:

 

f(-1) = 5

f(2) = 0

 

Then I need to find where is derivative = 0

 

f'(x) = 4 - t² - 2t²

 

Then I set that to zero and solve. However, I'm actually having some trouble with the algebra. Any help?

 

#55, p281

ANSWER: f(2^1/2) = 2, f(-1) = -3^1/2

Answer 1

Never mind Mia.  You did a great job sofar.

However allow me the comments:.

First of all f(-1) is not 5 but -3.

f(2) = 0....correct.

Now to search for any max or min you should build the first derivative of f(t):

f(t)^′ = 4-3t².

Setting f(t)′ = 0 4-3t² we determine the critical points which are

t= ±2/(3)^1/2 i.e. both +1.15 and -1.15

At these points t=1.15 and t=-1.15 you should have extreme values either max or min.

To find out we should execute the second derivative to get:

f(t)” =-6t 

Discussion of this result

We substitute both values of the critical numbers in this expression, we get

-6(1.15) which is a negative value, therefore at this t value of 1.15 a maximum.

At t=-1.15: If you plug in the expression -6t you get a positive value.  Therefore a minimum at this t value.

 

That means we have a minimum at t= -1.15 and a maximum at t=1.15

I hope this is helpful!

Answer 2

Your first derivative didn't use a correct form of differentiation.

 

f(t) = t(4-t²) = 4t - t³

 

f'(t) = 4 - 3t²

 

We set the derivative equal to zero

 

4 - 3t² = 0

 

4 = 3t²

 

4/3 = t²

 

t = +/- 2√3/3

 

When you do the first derivative test, you see that there is a relative minimum at t = -2√3/3 as well as a relative maximum at t = 2√3/3.

 

f(-2√3/3) = -2√3/3 * (4 - 12/9) = -16√3/9 = -3.08

f(2√3/3) = 16√3/9 = 3.08

f(-1) = -1(4-1) = -3

f(2) = 2(0) = 0

 

Out of all of these values, -3.08 is the lowest & 3.08 is the highest. Therefore an absolute maximum exist at t = 2√3/3 & an absolute minimum exists at t = -2√3/3.

Answer 3

When doing the algebra, you want to write the function in a way that will make it easier to derive.

 

f(x) = t(4 - t²)

 

      = 4t - t³

 

 

Now we use the power rule to derive each term.

 

f'(x) = 0

 

4 - 3t² = 0

 

 

-3t² = -4

 

  t² = 4 / 3

 

  t = ±√(4 / 3)

 

  t = - 2 / √3          and           t = 2 / √3

 

 

These are your critical points.  You can rationalize the denominator, but that is not necessary.  You can use your calculator to estimate those radicals.

 

Next, you need use test points and evaluate the derivative using those test points to see where you have a local maximum and local minimum.

 

I trust this part you can do.