f(x)= xsqrt((x^2)+16) is defined on the interval [-4,6] A. f(x) is concave down on the region ? B. f(x) is concave up on the region  ? C. The inflection point for this function is at x=?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`f(x)=x*sqrt(x^2+16)=x*(x^2+16)^(1/2)=>`

`f'(x)=1*(x^2+16)^(1/2)+x*1/2*(x^2+16)^(-1/2)*2x=>`

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

`f''(x)=1/2*(x^2+16)^(-1/2)*2x+[2x*sqrt(x^2+16)-x^2*1/2*(x^2+16)^(-1/2)*2x]/[x^2+16]=>`

`f''(x)=x/sqrt(x^2+16)+[2x*(x^2+16)-x^3]/[sqrt(x^2+16)*(x^2+16)]`

`f''(x)=[x*(x^2+16)+2x*(x^2+16)-x^3]/[sqrt(x^2+16)(x^2+16)]=>`

`f''(x)=[16x+2x^3+32x]/[sqrt(x^2+16)(x^2+16)]=>`

`f''(x)=[48x+2x^3]/[sqrt(x^2+16)(x^2+16)]`

Denominator is always positive =>

`f''(x)<0=>48x+2x^3>0=>`

`2x(24+x^2)>0=>2x>0=>x>0`

f(x) concave down when f''(x)<0=>concave down on `[-4,0)`

f(x) concave up when f''(x)>0=> concave up on `(0,6]`

Thus point of inflection at x=0

 

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team