Question

Find a number on the closed interva; [3/4,5/4] such that the sum of the number and its reciprocal is...
a.) as small as possible

b.) as large as possible

Answer 1

my answers are a.) 3/4 and b.) 5/4 but I am told these are wrong...

Answer 2

for a.) 3/4 is on the interval and if you add it to its reciprocal 4/3, you would get the smallest number possible, correct?

Answer 3

actually now that i checked with my calculator you can get values on the interval that when added to the reciprocal you get smaller numbers than (3/4)+(4/3)

Answer 4

Let f(x) = x + 1/x. Then take the first derivative and check to see if you can find if the first derivative is 0 inside the interval. If you find such an x, check to see if it is a min or max. If min, check against the f(endpoints) of the interval.

Answer 5

how did you know to use that equation

Answer 6

If you are adding a number and its reciprocal, the number is x and its reciprocal is 1/x. What you are doing is finding a value of a function in an interval, specifically, a closed interval, and it is an important distinction on whether the interval is open or closed. So, this is a function and it is the sum of what is indicated above.

Answer 7

ok, i have the first derivative, so i set that equal to zero and find x?

Answer 8

Even if the problem were something trivial like "find the largest value in the interval [10, 15], you could represent this by the function y=x, which is a trivially simple function.

Answer 9

If that is even possible. It might be the case that the first derivative is not 0 in that interval or even for all x (outside the interval), but it has to be checked for 0 within the interva, to see if we have any critical values.

Answer 10

ok critical value is x= +-1

Answer 11

Ok, good, but we are talking about a function that is defined (for our purposes) only in the specified interval. The function itself is valid over more than just our interval, but we have set the limitation of just this interval, because we can, and because it leads us to the solution. So, we are only concerned about +1 for x as a critical value. Now, you could take the second derivative and see that the second derivative is positive in that interval, or just realize that the first derivative goes from - to + around x=1. So we have a relative min for ALL x. And since we are talking only about the interval, you HAVE TO check f(x) at the endpoint of the interval against f(1).

Answer 12

So, your function looks like a bowl (concave up) with f(1) at the bottom of the bowl so that value of f(1) will be the min, so this boils down to checking f(3/4) and f(5/4) to see what will be the max.

Answer 13

ok the largest value is f(3/4)

Answer 14

and i found the second derivative which is 2/x^3 and the value is f''(1) is positive which means that 1 is the minimum, correct?

Answer 15

Actually, f(1) is the minimum which is equal to 2.

Answer 16

right, the value of f''(1) is positive

Answer 17

I've got a neat picture of the graph for you.

Answer 18

oh nice

Answer 19

That website in the corner is a free online graphing calculator and much more. And easy to use!

Answer 20

I just changed the scale on x and y for the graph, and you can see the max and min very clearly. It's a great tool. You might want to experiment with it.

Answer 21

thanks, i like the way it looks

Answer 22

i will thank you

Answer 23

I'm going to attach another shot of the graph. Hold on.

Answer 24

I'm sure I could do better with more familiarity but you get the picture.

Answer 25

much easier to see the max vs. min

Answer 26

Even though it could still be better, you better see the difference in f(x) values now.

Answer 27

Sorry, I worded that backwards. Trying again. Even though it could still be better, you CAN better see the difference in f(x) values now.

Answer 28

My first wording was awkward, sorry.

Answer 29

"better see" was meant to be typed "see better". Maybe I'm slightly dyslexic.

Answer 30

haha no worries

Answer 31

Any way. I hope I'm helping and I really appreciate the recognition!

Answer 32

yes thank you very much for the help

Answer 33

Good luck to you in all of your studies, and it was a pleasure working with you! I hope we run into each other again!

Answer 34

definitely, take care