Is #f(x)=(x^2+2x-6)/(2x+1)# increasing or decreasing at #x=0#?

Answer 1

#uarr#. Look for x = o, on the graph that is a hyperbola, with asynptotes #y =x/2+3/4 and 2x+1=0#.

By precise division,

#f=x/2+3/4-(27/4)/(2x+1)# and
#f'=1/2+(27/2)/(2x+1)^2=1/2+27/8=31/8>0#, at x = 0.
So, f #uarr#, at # (0, -6)#.

Examine the graph.

Take note that the graph has asynptotes and is a hyperbola.

#y =x/2+3/4 and 2x+1=0#. These meet at the
center #(-1/2, 1/2)#

graph{(x^2+(y+6)^2-.1)=0 [-20, 20, -10, 10]}

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Answer 2

To determine if ( f(x) = \frac{x^2 + 2x - 6}{2x + 1} ) is increasing or decreasing at ( x = 0 ), we can use the first derivative test.

Calculate the first derivative of ( f(x) ), denoted as ( f'(x) ). Then, substitute ( x = 0 ) into ( f'(x) ) and determine the sign of the result.

[ f'(x) = \frac{(2x + 1)(2x + 1) - (x^2 + 2x - 6)(2)}{(2x + 1)^2} ]

[ f'(x) = \frac{4x^2 + 4x + 1 - 2x^2 - 4x + 12}{(2x + 1)^2} ]

[ f'(x) = \frac{2x^2 + 13}{(2x + 1)^2} ]

Now, substitute ( x = 0 ) into ( f'(x) ):

[ f'(0) = \frac{13}{(0 + 1)^2} = 13 ]

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

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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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