Determine the value of the constant a if the function #f(x)# defined below is continuous at #x=2#. #f(x)={(ax^2+7x; x≤2),(3x^2+3a; x>2):}# ?

Answer 1

Please see below.

#f# is continuous at #a# if and only if #f(a) = lim_(xrarra)f(x)#.
In order for #f# to be continuous at #2#, we need #lim_(xrarr2)f(x)# to exist and to be equal to #f(2)#.
This function changes rules at #2#, so to make sure that the limit at #2# exists, we need the left and right limits at #2# to exist and to be equal.
So, find #lim_(xrarr2^-)f(x)# and find #lim_(xrarr2^+)f(x)#
Set those equal to each other and solve for #a#.
Check the #f(2)# gives the same value as the limit.
You should get #a = -2#
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Answer 2

#a = -2#

If the piecewise function is continuous at #x =2#, then the pieces must be the same value at #x =2#, and we can, therefore, set them equal to each other:

#a(2^2)+ 7(2) = 3(2^2)+ 3a#

#4a+ 14 = 12+ 3a#

#a = -2#

#f(x)={(-2x^2+7x; x≤2),(3x^2-6; x>2):}#

Here is a graph of #f(x)#:

Please observe that the graph is continuous.

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Answer 3

# -2#.

For the given fun. #f# to be continuous at #x=2#, we must have,
#lim_(x to 2-)f(x)=f(2)=lim_(x to 2+)f(x)............(square)#.
Now, as #x to 2-, x lt 2. :. f(x)=ax^2+7x#.
#:. lim_(x to 2-)f(x)=lim_(x to 2-)ax^2+7x=a*2^2+7*2#.
#:. lim_(x to 2-)f(x)=4a+14......................(square1)#.
Similarly, #lim_(x to 2+)3x^2+3a=3*2^2+3*a#.
#:. lim_(x to2+)f(x)=3a+12......................(square2)#.
Then, from #(square), (square1) and (square2)#, we have,
# 4a+14=3a+12," giving, "a=-2#.
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Answer 4

To determine the value of the constant ( a ) such that the function ( f(x) ) is continuous at ( x = 2 ), we need to ensure that the limit of the function as ( x ) approaches ( 2 ) from both sides is equal.

  1. First, find the expression for ( f(x) ) when ( x ) approaches ( 2 ) from the left side (denoted as ( f(2^-) )):

[ f(2^-) = a(2)^2 + 7(2) ]

  1. Next, find the expression for ( f(x) ) when ( x ) approaches ( 2 ) from the right side (denoted as ( f(2^+) )):

[ f(2^+) = 3(2)^2 + 3a ]

  1. Since the function is defined as continuous at ( x = 2 ), ( f(2^-) ) should equal ( f(2^+) ):

[ a(2)^2 + 7(2) = 3(2)^2 + 3a ]

[ 4a + 14 = 12 + 3a ]

[ 4a - 3a = 12 - 14 ]

[ a = -2 ]

So, the value of the constant ( a ) that makes the function ( f(x) ) continuous at ( x = 2 ) is ( a = -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.

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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