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...
The purpose of finding the max/min of 1/f(x) is to determine the maximum or minimum value of a given function, which can be useful in analyzing and understanding the behavior of the function.
The max/min of 1/f(x) can be calculated using various methods such as taking the derivative, setting the derivative equal to zero and solving for x, or using graphical methods such as finding the point of intersection of the tangent line with the x-axis. The specific method used will depend on the function and the available tools.
Yes, the max/min of 1/f(x) can be negative. This means that the function has a negative maximum or minimum value. It is important to consider the context of the function and the domain and range when interpreting a negative max/min value.
Finding the max/min of 1/f(x) has many real-world applications in fields such as physics, economics, and engineering. For example, it can be used to optimize production costs, determine the most efficient design for a structure, or predict the trajectory of a projectile.
There may be limitations to finding the max/min of 1/f(x) depending on the complexity of the function and the available tools. For some functions, it may not be possible to find an analytical solution and numerical methods may be required. Additionally, the max/min found may be an estimate rather than an exact value due to limitations in precision and accuracy.