How can I work this to find what X equals? It's a Cubic equation

Calculus
Tutor: None Selected Time limit: 1 Day

This is the first derivative from a quartic, I'm just trying to find the min and max

0.36x^3 - 0.84x^2 +5.52x - 6.84 = 0

I thought about making them all fractions? Like:

9x^3/25 - 21x^2/25 +138x/25 - 171/25 = 0

but I wasn't sure how to work it out from there

Apr 22nd, 2015

If this is indeed the problem (please recheck the data) , the second step is not necessary. You can start from 

0.36x^3 - 0.84x^2 +5.52x - 6.84 = 0

This equation is not amenable to a simple algebraic solution. It can be solved numerically. 

A simple and intuitive method to do it is the following:

You can also plot it with excel, For example substitute ten value in the range 1 to 10: 1, 2, 3...10. You'll find that the solution fall between 1 and 2 (you'll see the graph crosses the x axis). Than plot again (it's fairly easy) for 10 value between 1 an 2: 1.0, ,1.1., 1.2...1.9, 2.0. You' ll find the solution between 1.3 and 1.4. Repeat for values between 1.3 and 1.4/ You'll find it is very close 1.36. In fact it is 1.356...

This is the only zero (you can see by plotting for x<1 and x>10 the graph moves away from the x axis)

If you plot correctly you'll also discover that the the derivative 0.36x^3 - 0.84x^2 +5.52x - 6.84 is ZERO at 1.356. This means that the function has only an inflection point there  - in fact x =1.356 is neither a minimum nor a maximum. Your original function - the integral of 0.36x^3 - 0.84x^2 +5.52x - 6.84 does not have maxima or minima....

Apr 22nd, 2015

avi300, thank you for helping me out.  What I think I'm needing help on is the work that gets that equation down to it's result of x = 1.356

Can you show me how we get that?

Apr 22nd, 2015

I showed you one way by plotting. Another is the goal seek function in Excel, where you setup the calculation on a worksheet and require it to attain zero. It should find this value of x.

In any case - are you sure the initial equation is correct?

Apr 22nd, 2015
goal seek.xlsx
Goal seek on F12 by varying c7 gives the exact answer. You can set C7 to be 1 and see that it works


Apr 22nd, 2015

Yes, the initial equation is the first derivative from:

y = 0.0089x^4 - 0.28x^3 + 2.76x^2 - 6.84x + 21.95

I'm asked to find the min/max of the first derivative and the inflection point on the second derivative.  I did with the second one already, just stuck on the first because its a cubic and I'm not sure how to get to x = 1.356 

Apr 22nd, 2015

Let me recheck this one just to make sure...hold on a few minutes

Apr 22nd, 2015

You have a big error. The derivative is:

0.0356x^3-0.84x^2+5.52x-6.84

This changes things dramatically. You have now 3 roots: 9.93 (maximum), 1.6 (min) 12.05 (min)

You can get them by playing with goal seek (after first plotting the function)

Apr 22nd, 2015

Uh oh.  Dang, I didn't thinking rounding made that much of a difference.  Thank you for that!

I like goal seek a lot but I need to show work for how I get the roots

Apr 22nd, 2015

First by looking at the function (not the derivative) you can let x-> inf or x-> -inf and you see the function is symmetric (it goes in both sides to + inf) it means that it MUST have a minimum somewhere. (I did not have the original function at the beginning so I did not see it right away. 

Second step is plotting the derivative with Excel. I would suggest in the range -15 to +15.

Then search with goal seek on the derivative around suspected zeros you see in the plot/

The find the values of the 2nd derivative at the zeros: negative -> max, positive - > min

Apr 22nd, 2015

It goes in both directions to + inf is enough for a minimum (it is nearly symmetric)

Apr 22nd, 2015

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