How do you factor 9x^2+10x-3?
I can't seem to figure it out! The original question was to give the approximate location of a local maximum for the function f(x)=3x^3+5x^2-3x. Help please! Thank you!
Update:Please show me all the steps. Thanks!
I can't seem to figure it out! The original question was to give the approximate location of a local maximum for the function f(x)=3x^3+5x^2-3x. Help please! Thank you!
Update:Please show me all the steps. Thanks!
Comments
Ok, you taken the derivative correctly but you should have specified and interval too.
Setting the derivative = 0 and solving is your next step. Using the QF you get x = -1.357 rounded and x = 0.246 rounded. Now you have to use the 2nd derivative test to see if one of these is a local max.
f"(x)= 18x+10 and substituting in the 1st root you get f"(-1.357) is <0 so this root will give a local
max while f"(0.246) will be >0 and will give a local min.
Plugging in -1.357 into the original formula will give you approximately 5.78 rounded.
9x^2 + 10x - 3
You will need to use the quadratic equation to solve this.
x = (-10 ± â(10^2 - 4 * 9 * (-3)) / 2 * 9
= (- 10 ± â208) / 18
= (- 10 ± 4â13) / 18
= (-5 ± 2â13) / 9
Just so you know, here is the graph http://imgur.com/YXBvP
Also, if you would like to check your work, you can find the max/min with http://www.bcalc.net/
For future references: This is the quadratic formula for ax^2 + bx + c = 0
http://2.bp.blogspot.com/_MeBS83wN8bE/TTRfImJEe8I/...