How do you find the extrema for #g(x) = sqrt(x^2 + 2x + 5)#?
Note that:
So the function
But this is a second order polynomial with leading positive coefficient, hence it has no maximum and a single local minimum.
and:
and
Consequently:
and:
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We need

For
#x<1# we have#g'(x)<0# so#g# is strictly decreasing in#(oo,1]# 
For
#x># #1# we have#g'(x)>0# so#g# is strictly increasing in#[1,+oo)# Hence
#g(x)>=g(1)=2>0# ,#AA# #x# #in# #RR# As a result
#g# has a global minimum at#x_0=1# ,#g(1)=2#
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To find the extrema for ( g(x) = \sqrt{x^2 + 2x + 5} ), you first need to find its derivative. Then, set the derivative equal to zero and solve for ( x ). Finally, determine whether these critical points correspond to maximum or minimum values by using the second derivative test or analyzing the behavior of the function around those points.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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