Optimize Function: Abs Max & Min Values of -2x^2 + 3x + 6x^(2/3) + 2

In summary, the absolute maximum and absolute minimum values of the function f(x) = -2x^2 + 3x + 6x^(2/3) + 2 are 5.481 and 3, respectively. The derivative of the function is f'(x) = -4x + 3 + 4x^-1/3, and to find the value of x where f'(x) = 0, one can graph the function on a graphing calculator and observe the x-values where the y-value is zero. Additionally, the function f(x) is defined for all real numbers, so there are no numbers at which it does not exist, but the derivative is not defined at x =
  • #1
pyrosilver
39
0

Homework Statement



Find the abs max and abs min values of the function
f(x) = -2x^2 + 3x + 6x^(2/3) + 2.


Homework Equations





The Attempt at a Solution



So the candidates are the endpoints, where f'(x) = 0, and where f(x) DNE.
f(-1) = 3
f(3) = 5.481

For the derivative of the function, i got to f(x) = -4x + 3 + (1/(4x^3)). Set it to 0. But I can't find what x is. How do I use my calculator to find this? For the past five years, I haven't been allowed to use a calculator for my math courses, so I don't really know how to do anything with it. This year, I can use one, but I'm not sure how to use it to find x.

Also, f('x) DNE at x = 0. How do I use that information to determine the value to put that as a candidate for the abs max and abs min?
 
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  • #2
pyrosilver said:

Homework Statement



Find the abs max and abs min values of the function
f(x) = -2x^2 + 3x + 6x^(2/3) + 2.


Homework Equations





The Attempt at a Solution



So the candidates are the endpoints, where f'(x) = 0, and where f(x) DNE.
f(-1) = 3
f(3) = 5.481

For the derivative of the function, i got to f(x) = -4x + 3 + (1/(4x^3)).
Your derivative has two errors. It should be f'(x) = -4x + 3 + 4x-1/3. Note that the last term is the same as 4/x1/3, not 1/(4x3), as you have.
pyrosilver said:
Set it to 0. But I can't find what x is. How do I use my calculator to find this? For the past five years, I haven't been allowed to use a calculator for my math courses, so I don't really know how to do anything with it. This year, I can use one, but I'm not sure how to use it to find x.
Solving for the value that makes f'(x) = 0 involves solving a somewhat messy cubic equation. There are ways to do this analytically, but they're not usually taught.

It could be that you are expected to graph the derivative on your calculator and then observe the value(s) of x at which the y value is zero. I'm assuming you have a graphing calculator. If you don't know how to do this, look at the instruction manual for your calculator.
pyrosilver said:
Also, f('x) DNE at x = 0. How do I use that information to determine the value to put that as a candidate for the abs max and abs min?
 
  • #3
Ah okay, thanks I'll fix that up.

Actually I don't currently have a graphing calculator, but i just plotted it out on an online one. To make sure the graphing calculator wasn't faulty, is x = .799999? Sometimes the online ones are a bit sketchy.

Thanks for your help!

Anyone know about the DNE case?
 
  • #4
The value I'm getting where the derivative is zero is about 1.6.

The domain of f is all real numbers, so there are no numbers at which f does not exist. The derivative isn't defined at x = 0, but you already knew that.
 
  • #5
Ah okay, thank you Mark44!
 

Related to Optimize Function: Abs Max & Min Values of -2x^2 + 3x + 6x^(2/3) + 2

1. What is the purpose of optimizing a function?

The purpose of optimizing a function is to find the maximum or minimum value of the function. This can help to identify the most efficient or optimal solution in a given situation.

2. How can I find the abs max and min values of a function?

To find the absolute maximum and minimum values of a function, you can use calculus techniques such as taking the derivative and solving for critical points, or you can use graphical methods such as finding the x-intercepts or using a graphing calculator.

3. What is the significance of the -2x^2 term in the given function?

The -2x^2 term is the quadratic term in the function and represents the overall shape of the graph. It can affect the location and value of the maximum and minimum points.

4. Can the abs max and min values of a function change?

Yes, the absolute maximum and minimum values of a function can change if the function is altered or if the domain is restricted. Additionally, a function may have multiple maxima and minima, so the values can change depending on which point is being considered.

5. How can optimizing a function be applied in real-life situations?

Optimizing a function can be useful in a variety of real-life situations, such as maximizing profits in a business, minimizing costs in production, or finding the ideal conditions for a scientific experiment. It can also be used in fields like engineering, economics, and biology to find the most efficient or effective solutions.

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