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

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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
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....
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?
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?
Goal seek on F12 by varying c7 gives the exact answer. You can set C7 to be 1 and see that it works
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
Let me recheck this one just to make sure...hold on a few minutes
You have a big error. The derivative is:
0.0356x^30.84x^2+5.52x6.84
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)
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
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
It goes in both directions to + inf is enough for a minimum (it is nearly symmetric)
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