CALCULUS HELPPPP!!!!
Favorites|Homepage
Subscriptions | sitemap
HOME > Mathematics > CALCULUS HELPPPP!!!!

CALCULUS HELPPPP!!!!

[From: ] [author: ] [Date: 11-09-21] [Hit: ]
(x,Differentiate the function.to find the tangent line,and since the slope of a horizontal line = 0let y =0,since y is quadratic, you will have two solutions.......
- Can you explain to me how you got it cause i would like to understand (: Thanks!

Find the points on the curve y = 2x^3 + 3x^2 − 12x + 7where the tangent line is horizontal.

(x,y)= (_____) (smaller x-value)
(x,y)= (_____) (larger x-value)



Differentiate the function.
g(x) = x^2(1 − 5x)

-
y = 2x^3 + 3x^2 − 12x + 7

to find the tangent line, you need to find its slope by taking derivative of y

y' = 6x²+6x -12

and since the slope of a horizontal line = 0
let y' =0, solve for x
plug value of x back into y

since y' is quadratic, you will have two solutions.
1
keywords: HELPPPP,CALCULUS,CALCULUS HELPPPP!!!!
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .