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}#

Now use the Quotient Rule and Chain Rule to find the second derivative:

#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})#

So

#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

To determine whether the function ( f(x) = x \sqrt{x^2+2x+5} + \sqrt{x^2+2x+5} ) is concave up or concave down, we first find its second derivative, denoted as ( f''(x) ). Then, we analyze the sign of ( f''(x) ) to identify the concavity of the function and its intervals.

After computing the second derivative, we evaluate it at critical points and any other points of interest to determine where the function changes concavity. The critical points are found by setting the first derivative equal to zero and solving for ( x ).

If ( f''(x) > 0 ) for a particular interval, the function is concave up on that interval. If ( f''(x) < 0 ), the function is concave down on that interval.

In summary, to determine concavity and its intervals:

  1. Compute the second derivative ( f''(x) ).
  2. Find critical points by setting the first derivative equal to zero and solving for ( x ).
  3. Evaluate ( f''(x) ) at critical points and any other points of interest.
  4. Analyze the sign of ( f''(x) ):
    • If ( f''(x) > 0 ), the function is concave up on the interval.
    • If ( f''(x) < 0 ), the function is concave down on the interval.
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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.

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.

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.

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.

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