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

[pyrosilver]

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?

 

[Mark44]

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/x^1/3, not 1/(4x³), as you have.

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.

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?

 

[pyrosilver]

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?

 

[Mark44]

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.

 

[pyrosilver]

Ah okay, thank you Mark44!