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Sagot :
To solve for the constants [tex]\( A \)[/tex] and [tex]\( B \)[/tex] that make the function [tex]\( f(x) \)[/tex] continuous for all [tex]\( x \)[/tex], we need to ensure that the function's values and limits match at the point where the pieces come together, specifically at [tex]\( x = 1 \)[/tex].
1. Left-hand limit as [tex]\( x \)[/tex] approaches 1:
The expression for [tex]\( f(x) \)[/tex] when [tex]\( x < 1 \)[/tex] is [tex]\( A x^2 + 5 x - 9 \)[/tex].
We need to find the limit of this expression as [tex]\( x \)[/tex] approaches 1 from the left:
[tex]\[ \lim_{x \to 1^-} (A x^2 + 5 x - 9) \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ A(1)^2 + 5(1) - 9 = A + 5 - 9 = A - 4 \][/tex]
Hence, the left-hand limit as [tex]\( x \)[/tex] approaches 1 is:
[tex]\[ \lim_{x \to 1^-} f(x) = A - 4 \][/tex]
2. Right-hand limit as [tex]\( x \)[/tex] approaches 1:
The expression for [tex]\( f(x) \)[/tex] when [tex]\( x > 1 \)[/tex] is [tex]\((3-x)(A-2x)\)[/tex].
We need to find the limit of this expression as [tex]\( x \)[/tex] approaches 1 from the right:
[tex]\[ \lim_{x \to 1^+} (3 - x)(A - 2x) \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ (3 - 1)(A - 2(1)) = 2(A - 2) = 2A - 4 \][/tex]
Hence, the right-hand limit as [tex]\( x \)[/tex] approaches 1 is:
[tex]\[ \lim_{x \to 1^+} f(x) = 2A - 4 \][/tex]
3. Continuity at [tex]\( x = 1 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 1 \)[/tex], the left-hand limit and right-hand limit must be equal to the value of the function at [tex]\( x = 1 \)[/tex], which is [tex]\( B \)[/tex]:
[tex]\[ \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) = B \][/tex]
This gives us two equations:
[tex]\[ A - 4 = B \][/tex]
[tex]\[ 2A - 4 = B \][/tex]
4. Solving the system of equations:
We solve these two equations simultaneously. Let's start with the first equation:
[tex]\[ B = A - 4 \][/tex]
Substitute [tex]\( B = A - 4 \)[/tex] into the second equation:
[tex]\[ 2A - 4 = A - 4 \][/tex]
Simplify to solve for [tex]\( A \)[/tex]:
[tex]\[ 2A - 4 = A - 4 \][/tex]
[tex]\[ 2A - A = -4 + 4 \][/tex]
[tex]\[ A = 0 \][/tex]
5. Find [tex]\( B \)[/tex]:
Using [tex]\( A = 0 \)[/tex] in the equation [tex]\( B = A - 4 \)[/tex]:
[tex]\[ B = 0 - 4 = -4 \][/tex]
Therefore, the values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] that make the function [tex]\( f(x) \)[/tex] continuous for all [tex]\( x \)[/tex] are:
[tex]\[ A = 0 \][/tex]
[tex]\[ B = -4 \][/tex]
These values ensure that the function is continuous at [tex]\( x = 1 \)[/tex].
1. Left-hand limit as [tex]\( x \)[/tex] approaches 1:
The expression for [tex]\( f(x) \)[/tex] when [tex]\( x < 1 \)[/tex] is [tex]\( A x^2 + 5 x - 9 \)[/tex].
We need to find the limit of this expression as [tex]\( x \)[/tex] approaches 1 from the left:
[tex]\[ \lim_{x \to 1^-} (A x^2 + 5 x - 9) \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ A(1)^2 + 5(1) - 9 = A + 5 - 9 = A - 4 \][/tex]
Hence, the left-hand limit as [tex]\( x \)[/tex] approaches 1 is:
[tex]\[ \lim_{x \to 1^-} f(x) = A - 4 \][/tex]
2. Right-hand limit as [tex]\( x \)[/tex] approaches 1:
The expression for [tex]\( f(x) \)[/tex] when [tex]\( x > 1 \)[/tex] is [tex]\((3-x)(A-2x)\)[/tex].
We need to find the limit of this expression as [tex]\( x \)[/tex] approaches 1 from the right:
[tex]\[ \lim_{x \to 1^+} (3 - x)(A - 2x) \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ (3 - 1)(A - 2(1)) = 2(A - 2) = 2A - 4 \][/tex]
Hence, the right-hand limit as [tex]\( x \)[/tex] approaches 1 is:
[tex]\[ \lim_{x \to 1^+} f(x) = 2A - 4 \][/tex]
3. Continuity at [tex]\( x = 1 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 1 \)[/tex], the left-hand limit and right-hand limit must be equal to the value of the function at [tex]\( x = 1 \)[/tex], which is [tex]\( B \)[/tex]:
[tex]\[ \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) = B \][/tex]
This gives us two equations:
[tex]\[ A - 4 = B \][/tex]
[tex]\[ 2A - 4 = B \][/tex]
4. Solving the system of equations:
We solve these two equations simultaneously. Let's start with the first equation:
[tex]\[ B = A - 4 \][/tex]
Substitute [tex]\( B = A - 4 \)[/tex] into the second equation:
[tex]\[ 2A - 4 = A - 4 \][/tex]
Simplify to solve for [tex]\( A \)[/tex]:
[tex]\[ 2A - 4 = A - 4 \][/tex]
[tex]\[ 2A - A = -4 + 4 \][/tex]
[tex]\[ A = 0 \][/tex]
5. Find [tex]\( B \)[/tex]:
Using [tex]\( A = 0 \)[/tex] in the equation [tex]\( B = A - 4 \)[/tex]:
[tex]\[ B = 0 - 4 = -4 \][/tex]
Therefore, the values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] that make the function [tex]\( f(x) \)[/tex] continuous for all [tex]\( x \)[/tex] are:
[tex]\[ A = 0 \][/tex]
[tex]\[ B = -4 \][/tex]
These values ensure that the function is continuous at [tex]\( x = 1 \)[/tex].
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