need help again (quadratic functions)

abel muroi

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Jan 13, 2015
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I was given the problem g(x) = x5 - 8x3 + 20x and i was told to find the local maximum and minimum values of the function and the value of x at which each occurs

I am assuming that i have to begin by factoring the function right?

x5 - 8x3 + 20x
x(x4 - 8x2 + 20)
x(x2 - 4x + 5) (x2 - 4x + 4)

this is as far as i got, can anyone tell me how to find the "maximum" and "minimum" values of this function?
 
i double checked the function, the problem is still g(x) = x5 - 8x3 + 20x

i also double checked my work.. and (x2 - 4x + 5) times (x2 - 4x + 4) is not (x4 - 8x + 20)

but i am still confused as to how to factor this correctly

BTW i am trying to put the function into standard form.. by putting it into standard form, can i find the maximum and minimum values?
 
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I was given the problem g(x) = x5 - 8x3 + 20x and i was told to find the local maximum and minimum values of the function and the value of x at which each occurs

I am assuming that i have to begin by factoring the function right?

x5 - 8x3 + 20x
x(x4 - 8x2 + 20)
x(x2 - 4x + 5) (x2 - 4x + 4)

this is as far as i got, can anyone tell me how to find the "maximum" and "minimum" values of this function?

x5 - 8x3 + 20x

= x(x4 - 8x2 + 20)

substitute

u = x2

x4 - 8x2 + 20 = u2 - 8 u + 20 cannot be factored further in real domain
 
i double checked the function, the problem is still g(x) = x5 - 8x3 + 20x

i also double checked my work.. and (x2 - 4x + 5) times (x2 - 4x + 4) is not (x4 - 8x + 20)

but i am still confused as to how to factor this correctly

BTW i am trying to put the function into standard form.. by putting it into standard form, can i find the maximum and minimum values?
Abel, you really have to get the terminology down if you want precise answers. You can find the standard form of quadratic and that is f(x) = A(x-B)^2 + C. But what you gave us is a fifth degree polynomial. So what do you mean by standard form. Before you attempt to write something in a particular form you need to know what the form looks like.
I am not sure how to find the local max/min of a 5th degree equation with out using calculus. Are you allowed to use calculus. Do you know what a derivative is?
 
Abel, you really have to get the terminology down if you want precise answers. You can find the standard form of quadratic and that is f(x) = A(x-B)^2 + C. But what you gave us is a fifth degree polynomial. So what do you mean by standard form. Before you attempt to write something in a particular form you need to know what the form looks like.
I am not sure how to find the local max/min of a 5th degree equation with out using calculus. Are you allowed to use calculus. Do you know what a derivative is?

Sorry about that, it's a bit hard to use the correct terminology when i have a hard time understanding the concept in the first place.

And yes I am allowed to use calculus, but I don't think my professor has talked about this yet. So i figured i should be somewhat familiar with these fifth degree polynomials by asking here.

I'm not sure what a derivative is, but can you summarize it in a few words?
 
Sorry about that, it's a bit hard to use the correct terminology when i have a hard time understanding the concept in the first place.

And yes I am allowed to use calculus, but I don't think my professor has talked about this yet. So i figured i should be somewhat familiar with these fifth degree polynomials by asking here.

I'm not sure what a derivative is, but can you summarize it in a few words?
The derivative of f(x) is another function of x that gives the slope of tangent line of f(x) at any point on f(x).
 
I was given the problem g(x) = x5 - 8x3 + 20x and i was told to find the local maximum and minimum values of the function and the value of x at which each occurs

I am assuming that i have to begin by factoring the function right?

x5 - 8x3 + 20x
x(x4 - 8x2 + 20)
x(x2 - 4x + 5) (x2 - 4x + 4)

this is as far as i got, can anyone tell me how to find the "maximum" and "minimum" values of this function?

Using derivative:

f'(x) = 5x4 - 24x2 + 20 = 0 (for local extrema)

substitute u = x2 and we get

5u2 - 24u + 20 = 0 → u1,2 = [24 ± √(576-400)]/10

from above you can get to four extrema in real domain
 
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