Homework Help

[matagilb]

Hi first post here on free math help and i was hoping that i could get some assistance. I have the equation (1/12)x(x+6)(x-26) and i need help multiplying it out to get a 3rd degree polynomial. Also i need to find the first derivative of the equation so if somebody could tell me which derivative rule to use i would appreciate it. For the 3rd degree polynomial i got (1/12)x^3-(5/3)x^2-13x. Could somebody tell me if that's right and if not help me get going in the right direction. Thank you for your time and help.

 

[mmm4444bot]

I have the equation (1/12)x (x + 6) (x - 26)

That is not an equation; it's an expression.

Equations always contain an equal sign, like the following equation.

y = (1/12)x (x + 6) (x - 26)


i got (1/12)x^3 - (5/3)x^2 - 13x

This expansion is correct.


Also i need to find the first derivative of the [expression] so if somebody could tell me which derivative rule to use i would appreciate it.

Use the Power Rule, to derivate each of the three terms in this polynomial, one by one.

 

[matagilb]

Re:

I have the equation (1/12)x (x + 6) (x - 26)

That is not an equation; it's an expression.

Equations always contain an equal sign, like the following equation.

y = (1/12)x (x + 6) (x - 26)


i got (1/12)x^3 - (5/3)x^2 - 13x

This expansion is correct.


Also i need to find the first derivative of the [expression] so if somebody could tell me which derivative rule to use i would appreciate it.

Use the Power Rule, to derivate each of the three terms in this polynomial, one by one.

Thank you for your help and sorry about the equation expression mixup. So using the power rule i have gotten (1/4x)^2-(10/3x)-13
1 is that correct and 2 is that all that needs to be done with it. Once again thank you

 

[mmm4444bot]

So using the power rule i have gotten (1/4x)^2 - (10/3x) - 13

The first term is not correct, as typed. Only x gets squared, not 1/4.

1/4 x^2 - 10/3 x - 13

If you want to put grouping symbols around the first two coefficients, then you need to write this:

(1/4)x^2 - (10/3)x - 13


is that all that needs to be done

Yes, unless there are additional instructions not posted.

 

[matagilb]

Ok thank you. Last thing how would i go about using the expression (1/4)x^2-(10/3)x-13 and use the first derivative test to find the max/min. I tried factoring it out but the -13 is confusing me. By the way thank you for your help previously and for checking my work. Confidence boost

 

[mmm4444bot]

If we draw a line tangent to the curve of function f at a local minimum or maximum point, then that line must be horizontal.

In other words, all tangent lines have slope zero at local minimums and maximums.

Well, the value of f`(x) is the slope of the tangent line to the graph of f(x) at any location in the domain.

You need to find the values of x that cause f`(x) to evaluate to zero. There will be either a local minimum or maximum coming out of function f at each of those values of x.

Once you know where the local extremes (minimums/maximums) are located, you get their values by evaluating f(x).

Is your class using graphing calculators?

 

[matagilb]

I understand the concept of the first derivative test but i dont know how to do it with the expression i have. How do i set (1/4)x^2-(10/3)x-13 to zero to find my answers? This might require a step by step im new to the derivative test.
Lets say f(x)=x^2-8x+4 i know that after finding the first derivative it becomes f'(x)=2x-8 and when setting it to zero i get x=4. So i dont want you to think that im just here for answers and nothing else. Im here for answers and to learn how to do it as well.

 

[matagilb]

Re:

Is your class using graphing calculators?

Yes

 

[matagilb]

Nevermind i figured it out i forgot about the quadratic formula. :mrgreen: Thank you for all your help though i appreciate it.

 

[mmm4444bot]

You posted that you did not know how to use the derivative; I misunderstood your intent.

You actually meant to ask, "How do I solve the following equation"?

1/4 x^2 - 10/3 x - 13 = 0

Nevermind

i forgot about the quadratic formula Gasp!

Heh, heh, heh. (Don't do that again.)

I asked about the graphing because once you have maximum and minimum values of f(x), you could identify each (max or min) easily by looking at the graph of f at those locations. This would save you the time of further investigation algebraically or using other tests.

Cheers ~ Mark