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

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pramodpandey | College Teacher | (Level 3) Valedictorian

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Let the function be

`f(x)=8x^2-(16x+4)ln(x)`  ,

Find the derivative of the function with respect to x.

`f'(x)=16x-16ln(x)-(16x+4)/x`

`=(16x^2-16xln(x)-16x-4)/x`

`=(16x^2-16(ln(x)+1)x-4)/x`

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

`16x^2-16(ln(x)+1)x-4=0`

`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`

`g(1.2)=.0301>0`

`g(1.1)=-0.54<0`

`g((1.2+1.1)/2)=g(1.15)=-1.332<0`

`g((1.15+1.25)/2)=g(1.175)=-.123<0`

`g((1.115+1.2)/2)=g(1.1875)=-0.047<0`

`g((1.1875+1.2)/2)=g(1.19375)=-.008<0`

`g((1.19375+1.2)/2)=g(1.196875)=0.01>0`

So root of equation may be approximately

`(1.196875+1.19375)/2=1.1953125`

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

`f''(x)=16-16/x+4/x^2`

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

x=1.1953125 is point of minima.

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

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

`f'(x)=16x-16ln(x)-(16x+4)/x`

`f'(x)=(16x^2-16xln(x)-16x-4)/x`

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

`16x^2-16xln(x)-16x-4=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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