Another question

L

Lizzie

Full Member
Joined
Sep 8, 2005
Messages
317
Consider the function y = f(x) = - x<sup>3</sup> + 9x + 8 on the interval [0, 4]. The minimum value of f(x) on the interval is: ?

How do I start this one? It reminds me of something from algebra, but I'd rather ask to make sure that I'm not doing things wrong.
 
Find the derivative.
Find where the derivative is zero.
Check the end points.
 
Ok

f'(x) = -3x<sup>2</sup> + 9

My calculator says when x=0, y=9 so..

f'(9) = 0 ??

and what do you mean by "check the end points"?
 
I think you're confusing derivatives with inverse functions.


You need to find WHERE f'(x)=0

f'(x)=-3x^2+9

-3x^2+9=0

x^2=3

x=root(3),-root(3)

Now, plot them on a number line. The endpoints are 0 and 4 (because this function is defined on [0,4]), so -root(3) can be thrown out. Thus, on a number line, you need to plot 0, root(3), and 4. Find the sign (positive or negative) of the derivative between 0 and root(3) and between root(3) and 4. If the sign changes from negative to positive at x=root(3), then f(root(3)) is a relative minimum.

Now, if it is a relative minimum (I haven't worked it out), you still aren't done, because you need to find the absolute minimum. The absolute minimum occurs at a relative minimum or at an endpoint. So you would need to find f(root(3)), f(0), and f(4) and see which is the least. That will be your answer.



EDIT: Naturally, I am incapable or basic math, and think that if x^2=3, then x=3,-3. Wow. I edited it. Should be right now.
 
Given f(x) = - x3 + 9x + 8 on the interval [0, 4], Find the minimum value of f(x) over the given interval.

Find f'(x) then find where

f'(x) is equal to zero (where the function MAY change from increasing to decreasing, or decreasing to increasing)

and where f'(x) = undefined or does not exist (at which point on this function does the derivative not exist? There are none on this function since the function is continous and differentiable over the entire interval [0,4].)

There is an error when you solved f'(x) = 0,

f'(x)=0 is equivalent to -3x^2+9=0, solving for x,

-3x^2=-9

3x^2=9

x^2 = 9/3

x = sqrt(1/3), -sqrt(1/3)

So the possible points where f(x) could have a relative minimum is where
x = 0, sqrt(1/3), 4 (we can remove -sqrt(1/3) because it is not in the interval [0,4].)

which are the points
(0,f(0)) , (sqrt(1/3),f(sqrt(1/3)) , (4,f(4))

Now evaluate f(x) for each value of x that we found to make f'(x) = 0, and
for each of the endpoints (because it may be possible for the endpoints to
be the minimum of the function on that interval). The point (x,f(x)) which
yields the smallest value for f(x) is your relative minimum.

Which point do you find for your relative minimum?

Have you covered using the first or second derivative test yet?

Correction: sqrt(1/3) and -sqrt(1/3) should be sqrt(3) and -sqrt(3). I made the mistake of reducing 9/3 to 1/3. My apologies. Hope this didn't too much cause much confusion. Thanks for catching that Gene
 
You are working your calculator backwards. You first want to solve f'(x)=0 or
-3x²+9=0
You should get two answers.
To check them (if you have a TI calculator) you could sto-> one of your answers in X then write
-3x²+9
hit enter and you should get 0 if you are right. Do the same with the second answer if both are in the range.

About "check the end points." You are interested in [0,4]. f'(x) only tells you about peaks and valleys. It is possible that say f(4) is lower than any valley but is not at valley itself. That would make it a minimum in the range.
 
Thanks to everyone for your help... This is a multiple choice question and the only answer that seems plausible is f(1), because the others are "less than -40", "-20", or "Is impossible to determine"

And would the maximum be sqrt 3?
 
Lizzie, Lizzie, Lizzie!
Opti gave three numbers to check.
x = 0, sqrt(1/3), 4
though he/she goofed on
x = sqrt(1/3) which should have been sqrt(3) which is, as you said a maximum so can't be a minimum. That leaves
x = 0, 4
to check.

Consider f(x) = - x3 + 9x + 8
Look at f(0) and f(4). Which of those is the smallest? That is the minimum. You don't even need a calculator for that (though it would be good calculator practice to use one to check your answers.)
 
One last question about this, I swear, and it's just checking. I know that the minimum is -20 and the maximum is sqrt3, thanks to everyone's help. Now, at x=0...is that a point of inflection?
 
You're correct Lizzie, because at the point (0,8) the function
behavior changes from concave up to concave down.
 
Yes. f''(x) = -6x = 0 gives the inflexion point at x=0. The curve goes down till that x = -sqrt(3) we thru out then starts up again as x gets more negative.
 
You must log in or register to reply here.
Top