Finding the Max/Min of 1/f(x): A Proof

[Michael_Light]

Homework Statement

If f(x)=2(x-5/4)²+15/8, then what is the maximum/minimum point of 1/f(x) and how can you prove it? Thanks...

Homework Equations

The Attempt at a Solution

I really have no ideas about this...

 

[Mentallic]

Well, if you were asked to find the turning point of some function g(x), then you would solve g'(x)=0. That is, you took the derivative of g(x) and equated it to 0.

So you want to find the turning point of 1/f(x). Again, you just need to differentiate it and equate it to 0. Can you find \frac{d}{dx}\left(\frac{1}{f(x)}\right) ?

 

[HallsofIvy]

Also, a clarification. Are you asked to find the max/min of 1/f(x) or f^{-1}(x)?

 

[Michael_Light]

Well, if you were asked to find the turning point of some function g(x), then you would solve g'(x)=0. That is, you took the derivative of g(x) and equated it to 0.

So you want to find the turning point of 1/f(x). Again, you just need to differentiate it and equate it to 0. Can you find \frac{d}{dx}\left(\frac{1}{f(x)}\right) ?

Isn't that will lead to a very complicated equation? Will it possible to have a easier method since the answer provided is that the function has a maximum value of 8/15 when x=5/4... which seems it is possible for us to just deduce the answer from the given equation without doing any calculation...

Also, a clarification. Are you asked to find the max/min of 1/f(x) or f^{-1}(x)?

I am asked to find 1/f(x)...

 

[Mentallic]

Isn't that will lead to a very complicated equation? Will it possible to have a easier method since the answer provided is that the function has a maximum value of 8/15 when x=5/4... which seems it is possible for us to just deduce the answer from the given equation without doing any calculation...

Well yes there is an easier way to deduce the answer based on logical deductions, but you just asked that you want to prove it, so you should do it the correct way.

Try answer it first and then see if it's that complicated or not.

 

[Mark44]

There's a simpler way than using the derivative as was already suggested. The graph of 2(x-5/4)²+15/8 is a parabola. You should be able to determine whether this parabola opens upward or downward, and should be able to find its vertex. After sketching the graph of y = f(x), sketch the graph y = 1/f(x). Each y value on this graph is the reciprocal of the corresponding y value on the graph of the parabola.

 

[Mentallic]

There's a simpler way than using the derivative as was already suggested. The graph of 2(x-5/4)²+15/8 is a parabola. You should be able to determine whether this parabola opens upward or downward, and should be able to find its vertex. After sketching the graph of y = f(x), sketch the graph y = 1/f(x). Each y value on this graph is the reciprocal of the corresponding y value on the graph of the parabola.

While I agree there is an easier way than taking the derivative, but if you had to prove it wouldn't taking the derivative be the correct approach?

 

[osnarf]

There's no one correct way to form a proof about something. Proving something just means you have shown that it must be true (or false, depending). For proving a max/min value, the derivative is reliable, but in many cases there will be another way.

For instance, if I wanted to prove to you that a catapult I built can throw a rock 50 feet in my backyard against the wind, we could sit there and start calculating all sorts of crazy things. Or we could just do the simplest thing, and shoot it and see if it goes further. It might go 200 feet, in which case it clearly can throw it 50 feet. If it throws the rock 45 feet then we might want to resort to something more complicated to see how to maximize it.

 

[Mentallic]

There's a simpler way than using the derivative as was already suggested. The graph of 2(x-5/4)²+15/8 is a parabola. You should be able to determine whether this parabola opens upward or downward, and should be able to find its vertex. After sketching the graph of y = f(x), sketch the graph y = 1/f(x). Each y value on this graph is the reciprocal of the corresponding y value on the graph of the parabola.

There's no one correct way to form a proof about something. Proving something just means you have shown that it must be true (or false, depending). For proving a max/min value, the derivative is reliable, but in many cases there will be another way.

Definitely, but even though it's blatantly obvious, I would like to be more rigorous than simply using the fact that the parabola has a minimum, thus its reciprocal has a maximum at the same x-value.

 

[Mark44]

While I agree there is an easier way than taking the derivative, but if you had to prove it wouldn't taking the derivative be the correct approach?

Not in a precalculus class where students had no knowledge of differentiation. Since this was posted in the Precalc section, it's reasonable to assume that the OP is not expected to use anything beyond precalculus.

 

[Mentallic]

Not in a precalculus class where students had no knowledge of differentiation. Since this was posted in the Precalc section, it's reasonable to assume that the OP is not expected to use anything beyond precalculus.

People often find their way into the wrong forum section, and if the OP instead replied with "I don't know calculus" or "we aren't supposed to use calculus" then we would be sure where to go from there.
I only pursued this approach because it is a definite possibility that this is the expected way to answer the question. I also want the OP to realize that it doesn't lead to a complicated equation...

Isn't that will lead to a very complicated equation?

 

[Mark44]

People often find their way into the wrong forum section

They certainly do!

, and if the OP instead replied with "I don't know calculus" or "we aren't supposed to use calculus" then we would be sure where to go from there.
I only pursued this approach because it is a definite possibility that this is the expected way to answer the question. I also want the OP to realize that it doesn't lead to a complicated equation...