Is #f(x)=sqrt(x+2) # increasing or decreasing at #x=2 #?

Answer 1

Cameron Alexander

#f(x)=sqrt(x+2# is increasing at #x=2#

To find whether #f(x)=sqrt(x+2# is increasing or decreasing at #x=2#, first differentiate #f(x)#

As #f(x)=sqrt(x+2# can be written as #f(x)=(x+2)^(1/2)#, hence

#(df)/(dx)=1/2(x+2)^(1/2-1)=1/2(x+2)^(-1/2)=1/(2sqrt(x+2)#

At #x=2#,

#(df)/(dx)=1/(2sqrt(2+2))=1/4#

As it is positive, #f(x)=sqrt(x+2# is increasing at #x=2#

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Answer 2

Christopher Allers

To determine if ( f(x) = \sqrt{x + 2} ) is increasing or decreasing at ( x = 2 ), we evaluate the derivative of ( f(x) ) at ( x = 2 ).

The derivative of ( f(x) ) with respect to ( x ) is: [ f'(x) = \frac{1}{2\sqrt{x + 2}} ]

Evaluating this derivative at ( x = 2 ): [ f'(2) = \frac{1}{2\sqrt{2 + 2}} = \frac{1}{2\sqrt{4}} = \frac{1}{4} ]

Since ( f'(2) > 0 ), the function is increasing at ( x = 2 ).

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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.