Finding max. and min. values analytically

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In summary, the conversation discusses finding the local minima and maxima of the function y = 2x^2-14x. The first step is to differentiate the function to find its derivative. Then, the critical point is found by setting the derivative equal to 0 and solving for x. The second derivative can also be used to determine if the critical point is a local minimum or maximum. Alternatively, noticing that the graph of the function is a parabola can also help in finding the critical point.
  • #1
lp27
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Homework Statement


y = 2x^2-14x


Homework Equations





The Attempt at a Solution


y' = 4x - 14
4x - 14 = 0
x = 14/4
f(14/4) = 2(14/4)^2 - 14 (14/4) = -49/2

I don't know how to continue from there. ((14/4), (-49/2)) is my critical point?
 
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  • #2
lp27 said:

Homework Statement


y = 2x^2-14x

Homework Equations


The Attempt at a Solution


y' = 4x - 14
4x - 14 = 0
x = 14/4
f(14/4) = 2(14/4)^2 - 14 (14/4) = -49/2

I don't know how to continue from there. ((14/4), (-49/2)) is my critical point?
there are a couple of steps involved with finding local min/max's:

1) differentiate
2) find f'(x) = 0
3) take values larger and smaller than x | f'(x) = 0 to see if it is a min/max [note: you must do this because you may have f'(x) = 0 without being a min/max]
min: \ /
max : / \
meaning, values less than x | f'(x) = 0 will have negative derivatives if it is a minimum value.. et c .
 
Last edited:
  • #3
You can also use the second derivative to determine a local maximum or local minimum. At any critical point c, if f''(c) > 0, there is a local minimum at (c, f(c)). If f''(c) < 0, there is a local maximum at (c, f(c)).

For your particular function, you could also notice that its graph is a parabola that opens upward. The only critical point is at the vertex of the parabola.

BTW you should simplify 14/4 to 7/2.
 

Related to Finding max. and min. values analytically

1. How do you find the maximum and minimum values analytically?

To find the maximum and minimum values analytically, you need to take the derivative of the function and set it equal to zero. Then, solve for the variable and plug it back into the original function to find the corresponding y-value.

2. What is the benefit of finding max. and min. values analytically?

Finding max. and min. values analytically allows for a precise and accurate determination of the extreme points of a function. This information can be useful in optimization problems or in understanding the overall behavior of a function.

3. Can you find max. and min. values analytically for any function?

Yes, you can find max. and min. values analytically for any function that is differentiable on the given interval. However, the process may be more complex for some functions and may require advanced calculus techniques.

4. Is there an alternative method to finding max. and min. values analytically?

Yes, you can also use graphical methods such as finding the critical points and using the first or second derivative tests to determine if they are maximum or minimum points on the graph.

5. Can the maximum and minimum values of a function occur at the same point?

Yes, it is possible for a function to have a maximum and minimum value at the same point. This occurs when the function has a horizontal point of inflection at that point. It is important to consider the second derivative in these cases to determine the nature of the point.

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