Question on derivative, maxima and minima

Sep 2019
12
1
England
Hi,

I'm having difficulty working out the maxima and minima of a number of easy equations, though I am grasping it quite well I'm still going wrong somewhere with the arithmetic, particularly where I'm trying to factor out the value(s) for x, after I have the derivative...

These are the two questions I am getting wrong so far...


(1) y = x + 1/x

y = x + x^-1

dy/dx = 1 - x^-2 = 0

1 = 1/x^2

x^2 = 1/1

x = square of 1

x = 1

d^2/dx^2 = 2x^-3

2(1)^-3 = 2

Minima

y = 2 + 1/2 = 2.5

My answer: MIN (2, 2.5)

But actual answer: MIN (1, 2) MAX (-1, -2)



(2) y = x^4 - 2x^2 + 1

dy/dx = 4x^3 - 4x = 0

4x(x - 1)(X + 1) = 0

x = 1 and -1

(here I think this is wrong)

d^2y/dx^2 = 12x^2 - 4

12(1)^2 - 4 = 8 Minimum
12(-1)^2 -4 = 8 Minimum

y = (1)^4 - 2(1)^2 + 1 = 1 - 2 + 1 = 0
y = (-1)^4 - 2(-1)^2 + 1 = 1 - 2 + 1 = 0

My answer: MIN (-1, 0) and MIN (1, 0) !!

Actual answer: MAX (0, 1), MIN (+/-1, 0)

It should be obvious where I've gone wrong but I'm not seeing it. I'd appreciate any help in setting me right. Thanks.
 
Sep 2019
12
1
England
correction for question (1)

y = 1 + 1/1 = 2

My answer: MIN (1, 2)

Actual answer: MIN (1, 2) MAX (-1, -2)
 

Plato

MHF Helper
Aug 2006
22,470
8,640
(1) y = x + 1/x
But actual answer: MIN (1, 2) MAX (-1, -2)
(2) y = x^4 - 2x^2 + 1
My answer: MIN (-1, 0) and MIN (1, 0) !!
Actual answer: MAX (0, 1), MIN (+/-1, 0)
Your real mistake is not posting the actual question that you are asked to answer,
You may well think that you did, but you did not!
You posted a lot of your work thinking we can put it together to see what the question was.
It simply does not work that way.
If \(\displaystyle y=x+\frac{1}{x}\) the first step is to look at its graph
After looking at its graph it is clear that there is no absolute maximum nor minimum.
There are relative max & min.
\(\displaystyle D_x\left(x+\frac{1}{x}\right)=\left(1-\frac{1}{x^2}\right)\)
Now it is clear that the derivative is zero at \(\displaystyle x=\pm 1\)
Look at the graph; which is which?
Can you use the second derivative to explain which is which?
 
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Debsta

MHF Helper
Oct 2009
1,298
592
Brisbane
Hi,

I'm having difficulty working out the maxima and minima of a number of easy equations, though I am grasping it quite well I'm still going wrong somewhere with the arithmetic, particularly where I'm trying to factor out the value(s) for x, after I have the derivative...

These are the two questions I am getting wrong so far...


(1) y = x + 1/x

y = x + x^-1

dy/dx = 1 - x^-2 = 0

1 = 1/x^2

x^2 = 1/1

x = square of 1

x = 1

d^2/dx^2 = 2x^-3

2(1)^-3 = 2

Minima

y = 2 + 1/2 = 2.5

My answer: MIN (2, 2.5)

But actual answer: MIN (1, 2) MAX (-1, -2)



(2) y = x^4 - 2x^2 + 1

dy/dx = 4x^3 - 4x = 0

4x(x - 1)(X + 1) = 0

x = 1 and -1

(here I think this is wrong)

d^2y/dx^2 = 12x^2 - 4

12(1)^2 - 4 = 8 Minimum
12(-1)^2 -4 = 8 Minimum

y = (1)^4 - 2(1)^2 + 1 = 1 - 2 + 1 = 0
y = (-1)^4 - 2(-1)^2 + 1 = 1 - 2 + 1 = 0

My answer: MIN (-1, 0) and MIN (1, 0) !!

Actual answer: MAX (0, 1), MIN (+/-1, 0)

It should be obvious where I've gone wrong but I'm not seeing it. I'd appreciate any help in setting me right. Thanks.
For Q1.
Note that if x^2 -1, the x=1 OR x=-1. You've neglected the second solution.
For Q2.
If 4x(x - 1)(x + 1) = 0, then x=0 OR x=1 OR x=-1. You've neglected 0 solution.
 
Mar 2012
564
29
I think your first mistake started here as Debsta said

x^2 = 1/1

x = square of 1

x = 1

\(\displaystyle
x = \pm\sqrt{1} = \pm1
\)

so this tells you that you have two critical numbers. figuring they are maxima or minima is the easy part
 
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Sep 2019
12
1
England
Your real mistake is not posting the actual question that you are asked to answer,
You may well think that you did, but you did not!
You posted a lot of your work thinking we can put it together to see what the question was.
It simply does not work that way.
If \(\displaystyle y=x+\frac{1}{x}\) the first step is to look at its graph
After looking at its graph it is clear that there is no absolute maximum nor minimum.
There are relative max & min.
\(\displaystyle D_x\left(x+\frac{1}{x}\right)=\left(1-\frac{1}{x^2}\right)\)
Now it is clear that the derivative is zero at \(\displaystyle x=\pm 1\)
Look at the graph; which is which?
Can you use the second derivative to explain which is which?
I am only going by what what my textbook has introduced me to, this is the section in full and how to deduce the maxima and minima from the second derivative. The questions I had gotten wrong are questions 8, 9 and 10 here. Also, it does not ask me to simply look at a graph, that would be too easy (or rather time consuming if I am drawing them all myself and not using a program)...

Max-Min.jpg

The graph for y = x + 1/x

Max-Min-2.jpg

So to the answer is clearly as I had given "actual answer Mininum at (1, 2) and Maximum at (-1. -2), But my question was why did I come to the wrong answer using the process given above in "example 4"? (and not from looking at a graph).
 

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Last edited:
Sep 2019
12
1
England
OK thanks
For Q1.
Note that if x^2 -1, the x=1 OR x=-1. You've neglected the second solution.
For Q2.
If 4x(x - 1)(x + 1) = 0, then x=0 OR x=1 OR x=-1. You've neglected 0 solution.
OK thanks, I can see now where I was going wrong :)
 
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