What are the critical point of this function 8*x^2-(16*x + 4)*ln(X)?

pramodpandey | Student

Let the function be

`f(x)=8x^2-(16x+4)ln(x)`  ,

Find the derivative of the function with respect to x.




To obtain citical points ,we need to equate `f'(x)=0`  ,and solve for x .


`4x^2-4(ln(x)+1))-1=0`               (i)

we will solve  (i) , by method of bisection.

let `g(x)=4x^2-4(ln(x)+1)-1`








So root of equation may be approximately


Tocheck max or min ,we need to find second derivative.


`{f''(x)}_{x=1.1953125}=5.41 >`

x=1.1953125 is point of minima.

pramodpandey | Student

let f(x)=8x^2-(16x+4)lnx



For critical points `f'(x)=0`


`4x^2-4xln(x)-4x-1=0`   (i)

Solving equation (i) ,by bisection method

we have x=1.65234375

which is critical point.

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