How do you determine whether the function #f (x) = x sqrt(x^2+2x+5)+1 sqrt(x^2+2x+5)# is concave up or concave down and its intervals?

Answer 1

It's concave up for #x > -1# (on the interval #(-1,infty)#) and concave down for #x < -1# (on the interval #(-infty,-1)#).

Write the function as #f(x)=(x+1)sqrt{x^2+2x+5}# and then use the Product Rule and the Chain Rule to get the first derivative:

#f'(x)=sqrt{x^2+2x+5}+(x+1) * 1/2 * (x^2+2x+5)^{-1/2} * (2x+2)#

#=(x^2+2x+5+x^2+2x+1)/sqrt{x^2+2x+5}=(2x^2+4x+6)/sqrt{x^2+2x+5}#

#f''(x)#

#=(sqrt{x^2+2x+5} * (4x+4)-(2x^2+4x+6) * 1/2 * (x^2+2x+5)^{-1/2} * (2x+2))/(x^2+2x+5)#

#=((x^2+2x+5)(4x+4)-(2x^2+4x+6)(x+1))/((x^2+2x+5)^{3/2})#

#=(4x^3+8x^2+20x+4x^2+8x+20-2x^3-4x^2-6x-2x^2-4x-6)/((x^2+2x+5)^{3/2})#

#f''(x)=(2x^3+6x^2+18x+14)/((x^2+2x+5)^{3/2})#

Since #x^2+2x+5>0# for all real numbers #x#, the sign (positive or negative) of #f''(x)# is determined by the sign of #2x^3+6x^2+18x+14#, which is determined by the sign of #x^3+3x^2+9x+7#.

The number #x=-1# is a real root of #x^3+3x^2+9x+7# since #-1+3-9+7=0#. If you use long-division of polynomials (or synthetic division), you'll find that #x^3+3x^2+9x+7=(x+1)(x^2+2x+7)#. The quadratic #x^2+2x+7# is always positive for all real #x#.

All of this implies that the sign of #f''(x)# is determined by the sign of #x+1#. Since #x+1>0# when #x > -1# and #x+1<0# when #x < -1#, the same is true for #f''(x)#.

This means the graph of #f(x)# is concave up for #x > -1# (on the interval #(-1,infty)#) and concave down for #x < -1# (on the interval #(-infty,-1)#).

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Answer 2

Paisley Johnson

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.