Why Does f(x) = 1/x Have a Minimum on [1,2] but Not on [-1,2]?

[domyy]

Homework Statement

Explain why f(x) = 1/x must have a minimum on [1,2] but not on [-1,2].

Homework Equations

The Attempt at a Solution

I don't know what to do. First, I took the x values from the given points and plugged them back into the function but the results do not coincide with the y values. I don't really know what to do.

I also thought about differentiating it and then f'(x) = 0.
I used the quotient rule:
(x)(1)' - (1)(x)'/(x)^2
-1/x^2 -> f'(x)=0
-1/x^2= 0

 

[vela]

[1,2] is the interval from 1 to 2, not the point (1,2).

 

[domyy]

I still don't know how to find the answer.
I've had exercises where I was able to write my answer as "MIN: (-1,-4) and (2,-4)" or "MIN: (0,0) and (3,0)" for instance and they were correct.

 

[Mark44]

I also thought about differentiating it and then f'(x) = 0.
I used the quotient rule:
(x)(1)' - (1)(x)'/(x)^2
-1/x^2 -> f'(x)=0
-1/x^2= 0

So what does this tell you?

One of the first things you should have done is sketch a graph of the function f(x) = 1/x. You would be able to see from the graph why there is a minimum on the interval [1, 2] and why there isn't one on the interval [-1, 2].

 

[domyy]

what I don't understand is how would the graph of [1,2] and [-1,2] differ from [-1,-4] and [2,-4], for instance, as I found in one of my previous homework.

 

[Mark44]

what I don't understand is how would the graph of [1,2] and [-1,2] differ from [-1,-4] and [2,-4], for instance, as I found in one of my previous homework.

This makes no sense. You aren't graphiing [1, 2] etc. These are intervals.

You didn't answer my previous question. You found that if f(x) = 1/x, then f'(x) = -1/x². Then you set f'(x) to 0, getting
-1/x² = 0

What is the solution to this equation?
What does that tell you about the graph of f(x)?
Have you graphed f(x) = 1/x?

 

[domyy]

there's no solution.

 

[Mark44]

What is the solution to this equation?

there's no solution.

Right. So what about the other questions I asked...

What does that tell you about the graph of f(x)?
Have you graphed f(x) = 1/x?

 

[HallsofIvy]

The crucial point is that f(0) is not defined. That matters on [-1, 2] but not on [1, 2]. Do you see why?

There is a theorem that says if a function is continuous on a closed and bounded interval it has both maximum and minimum values on the interval. Do you see why that applies to [1, 2] but not to [-1, 2]?

Finally there is a theorem that says that if a function is continuous on a closed and bounded interval and differentiable in its interior, then the max and min occur where the derivative is 0 or at the end points.