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labwork [276]
3 years ago
10

find the value of "a" and "b" for which the limit exists both as x approaches 1 and as x approaches 2:

Mathematics
1 answer:
lbvjy [14]3 years ago
3 0

Answer:

a = 4

b = -2

Step-by-step explanation:

If the given function is continuous at x = 1

\lim_{x \to 1^{-}} f(x)=(x+1)

                     =2

\lim_{x \to 1^{+}} f(x)=ax+b

                     =a+b

\lim_{x \to 1} f(x)=ax+b

                   =a+b

And for the continuity of the function at x = 1,

\lim_{x \to 1^{-}} f(x)=\lim_{x \to 1^{+}} f(x)=\lim_{x \to 1} f(x)

Therefore, (a + b) = 2 -------(1)

If the function 'f' is continuous at x = 2,

\lim_{x \to 2^{-}} f(x)=ax+b

                     =2a+b

\lim_{x \to 2^{+}} f(x)=3x

                     =6

\lim_{x \to 2} f(x)=3x

                   =6

Therefore, \lim_{x \to 2^{-}} f(x)=\lim_{x \to 2^{+}} f(x)=\lim_{x \to 2} f(x)

2a + b = 6 -----(2)

Subtract equation (1) from (2),

(2a + b) - (a + b) = 6 - 2

a = 4

From equation (1),

4 + b = 2

b = -2

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