Home > Calculus calculators > Find maximum and minimum value of y calculator

Solve any problem
(step by step solutions)
Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution
Find maximum and minimum value of y=x^3-9x^2+24x+2

Solution:
Your problem `->` maximum and minimum value of y=x^3-9x^2+24x+2


Here, `y=x^3-9x^2+24x+2`

`:. (dy)/(dx)=``=(x^(3)-9x^(2)+24x+2)`

`=x^(3)-9x^(2)+24x+2`

`d/(dx)(x^(3)-9x^(2)+24x+2)`

`=d/(dx)(x^(3))-d/(dx)(9x^(2))+d/(dx)(24x)+d/(dx)(2)`

`=3x^(2)-18x+24+0`

`=3x^(2)-18x+24`

`=3(x^(2)-6x+8)`

`=3((x-2)(x-4))`

`=3(x-2)(x-4)`

For stationary values, `(dy)/(dx)=0`

`=>3(x-2)(x-4)=0`

`=3(x^(2)-6x+8)`

`=>x=0`

`:.`At `x=0` we get stationary values.

Now, `(d^2y)/(dx^2)=``d/(dx)(3(x-2)(x-4))`

`=3(d/(dx)(x-2))(x-4)+3(x-2)(d/(dx)(x-4))`

`d/(dx)(x-2)=1`
`d/(dx)(x-2)`

`=d/(dx)(x)-d/(dx)(2)`

`=1-0`

`=1`


`d/(dx)(x-4)=1`
`d/(dx)(x-4)`

`=d/(dx)(x)-d/(dx)(4)`

`=1-0`

`=1`


`=3(1)(x-4)+3(x-2)(1)`

`=3(x-4)+3(x-2)`

`=3x-12+3x-6`

`=6x-18`

`=6x(1-3/x)`

`=6(x-3)`

`=6x-18`

`((d^2y)/(dx^2))_(x=0)``=6(0)-18`

`=0-18`

`=-18` (negative)

`:.` At `x=0` the function is maximum

Now, `y=x^3-9x^2+24x+2`

`"putting " x=0`

`y_(max)``=(0)^(3)-9(0)^(2)+24(0)+2`

`=0-0+0+2`

`=2`






Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me
  
 

Share with your friends, if solutions are helpful to you.
 
Copyright © 2019. All rights reserved. Terms, Privacy