The original poster is tasked with finding the maximum or minimum point of the function 1/f(x), where f(x) is defined as 2(x-5/4)² + 15/8. The discussion revolves around the methods to prove this without directly calculating the derivative.
There is an ongoing exploration of different methods to approach the problem, with some participants advocating for the use of derivatives while others suggest visualizing the function's graph. The conversation reflects a mix of opinions on the best approach to prove the maximum or minimum without reaching a definitive consensus.
Participants note that the original poster may be expected to solve the problem within the constraints of precalculus, which raises questions about the appropriateness of using calculus-based methods.
Mentallic said: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 [tex]\frac{d}{dx}\left(\frac{1}{f(x)}\right)[/tex] ?
HallsofIvy said:Also, a clarification. Are you asked to find the max/min of 1/f(x) or [itex]f^{-1}(x)[/itex]?
Michael_Light said: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...
Mark44 said: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.
Mark44 said: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.
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.osnarf said: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.
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 said: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?
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.Mark44 said: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.
Michael_Light said:Isn't that will lead to a very complicated equation?
They certainly do!Mentallic said:People often find their way into the wrong forum section
Mentallic said:, 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...