The function f(x)=x*sqrt(x^2+3)+5 is concave up for x in the interval what?

Question

needing help please

Answer 1

To find the interval where f(x) is concave up, we take the second derivative and set it equal to zero.

 

Rewrite f(x) so that it is easier to derive.  We will use a the product rule and chain rule.  Be sure to check your algebra, so take your time and work this out.

 

f(x) = x(x² + 3)^1/2 + 5

 

 

d/dx(d/dx[f(x)]) = 0

 

d/dx[(x² + 3)^1/2 + x(1/2)(x² + 3)^-1/2 * 2x] = 0

 

d/dx[(x² + 3)^1/2 + x²(x² + 3)^-1/2] = 0

 

((1/2)(x² + 3)^-1/2 * 2x) + (2x(x² + 3)^-1/2 - x²(1/2)(x² + 3)^-3/2 * 2x) = 0

 

 

Next, we simplify the left side.

 

x(x² + 3)^-1/2 + 2x(x² + 3)^-1/2 - x³(x² + 3)^-3/2 = 0

 

 

Factor out the (x² + 3)^-1/2 term.  This is the GCF.

 

(x² + 3)^-1/2(x + 2x - x³(x² + 3)^-1) = 0

 

(x² + 3)^-1/2(3x - x³(x² + 3)^-1) = 0

 

 

Set the factors equal to zero.

 

1 / √(x² + 3) = 0           and        (3x(x² + 3) - x) / (x² + 3) = 0

 

                                                 3x³ + 8x = 0

 

                                                 x(3x² + 8) = 0

 

                                                 x = 0

 

x = 0 is our point of inflection.  The point of inflect is where f(x) changes concavity. 

 

 

Next, we perform test points around this point.  plug x = -1    and   x = 1 into the second derivative. 

 

If d/dx(d/dx(f(x)) is negative, it is concave down.

If d/dx(d/dx(f(x)) is positive , it is concave up.

 

I will let you complete this step on your own.

Answer 2

f'(x)= (2x^2+3)/sqrt(x^2+3)

 

f''(x)=[2x(x^2+9)]/(x^2+9)^(3/2)

 

f''(x)=0

 

So, the critical point is x=0

 

for x (0 - infinite).  it is concave up.

 

for x ( - infinite to 0) it is concave down.